Equational Logic A language L of algebras (or algebraic structures ) - - PDF document

(LMCS, p. 133) III.1

Equational Logic

A language L of algebras (or algebraic structures) consists of

• a set

F

• f function symbols

f, g, h, · · ·

• a set

C

• f constant symbols

c, d, e, · · ·

• a set

X

• f variables

x, y, z · · · . Each function symbol has an arity to indicate how many arguments it takes. If the symbol takes n arguments we say it is n-ary. n = 1 2 3 unary binary ternary

(LMCS, p. 133) III.2 Example The language LBA

• f Boolean algebras

has F = {∨, ∧,′ } C = {0, 1} ∨ and ∧ are binary function symbols,

′

is a unary function symbol.

symbol name

∨

join

∧

meet

′

complement

The constants are just called by the usual names zero and one.

(LMCS, p. 134) III.3 The meaning of the symbols Given a set A:

• Function symbols are interpreted as

functions on the set.

• Constant symbols are interpreted as

elements of the set. A function f that maps n-tuples of elements of A to A is called an n-ary function on A. We write f : An → A to say f is n-ary on A.

(LMCS, p. 134) III.4 We can describe a single unary function on a small set either with a table, e.g., f 1 1 2 3 3 3

• r with a directed graph representation:

1 2 3

(LMCS, p. 134) III.5 Cayley Tables We can also describe small binary functions

• n a set

A using a table, called a Cayley table. To describe the integers mod 4, with the binary operation of multiplication mod 4, we have the Cayley table: · 1 2 3 1 1 2 3 2 2 2 3 3 2 1

(LMCS, p. 135) III.6 Function Tables We can also describe functions on a small set A using a table that is similar to the truth tables used to describe the connectives. To describe the ternary function f(x, y, z) = 1 + xyz

• n the integers modulo 2

we could use x y z f 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

(LMCS, p. 135) III.7 Interpretations An interpretation I

• f the language

L

• n

a nonempty set A assigns to each symbol from L a function or constant as follows:

• I(c)

is an element of A for each constant symbol c in C.

• I(f)

is an n–ary function on A for each n–ary function symbol f in F.

(LMCS, p. 136) III.8 Visualizing an interpretation I

• n a set

A:

I(c) = c A I Symbol Interpretation f I I( f ) = x x

1

f

n

(LMCS, p. 135) III.9 Algebras An L-algebra (or L-structure)

A

is a pair (A, I) where I is an interpretation of L

• n

A. Given an algebra A: the interpretations of the constant symbols are called the constants of the algebra the interpretations of the function symbols are called the fundamental operations of the algebra.

(LMCS, p. 136) III.10 Notation We prefer to write cA for I(c) fA for I(f),

• r even simpler,

c for I(c) f for I(f), Also, rather than writing (A, I) we prefer to write (A, F, C)

• r to simply list the symbols in

F and C, e.g., the algebra (R, +, ·, 0, 1),

(LMCS, p. 137) III.11 Example Let L = LBA = {∨, ∧,′ , 0, 1} Let Su(U) be the collection of all subsets of a given set U (U is called the universe). Interpret join as union (∪) meet as intersection (∩) complement as complement (′) in U as the empty set (Ø) 1 as the universe (U). Then Su(U) = (Su(U), ∪, ∩,′ , Ø, U) is the Boolean algebra of subsets of U.

(LMCS, p. 138) III.12 Example Let L = LBA = {∨, ∧,′ , 0, 1} Let A = {0, 1}, and let the function symbols be interpreted as follows: ∨ 1 1 1 1 1 ∧ 1 1 1

′

1 1 and 0, 1 are interpreted in the obvious manner (as 0, 1). This is the best known of all the Boolean algebras.

(LMCS, p. 140) III.13 Terms are used to make the sides of equations. Examples of terms using familiar infix notation for the language {+, ·, −, 0, 1}: 1 x y −0 − 1 − x − y 1 + 0 x · y − (−x) x + 1 x · (y + z) (−x) · (−y) 1 + (0 + 1)

(LMCS, p. 140) III.14 Examples

• f terms using familiar infix notation for the

language of Boolean algebra {∨, ∧,′ , 0, 1}: 1 x y 0′ 1′ x′ y′ 1 ∨ 0 x ∧ y x′′ x ∨ 1 x ∧ (y ∨ z) (x′) ∧ (y′) 1 ∨ (0 ∨ 1)

(LMCS, p. 140) III.15 Examples

• f terms using prefix notation.

If f is a unary function symbol: x fx, ffx are terms. If c is a constant symbol then c fc ffc are terms. If g is a binary function symbol then gcx gyy ggfzcgcx are terms. If h is a ternary function then hxyz hccx hfgxcgxcggxyfc are terms.

(LMCS, p. 140) III.16 Terms The L-terms over X are (inductively) defined by the following:

• A variable x in

X is an L–term.

• A constant symbol c in C

is an L–term.

• If

t1, . . . , tn are L–terms and f is an n–ary function symbol in F, then ft1 · · · tn is an L–term.

(LMCS, p. 141) III.17 A Parsing Algorithm (for terms in prefix form) Define γ

• n the symbols of a string

s = fs1 · · · sn by:

• γ

at the first symbol f is 0.

• Increase

γ by one on variables and constants.

• Decrease

γ by arity − 1

• n the

remaining function symbols.

f x c g γ = 0 γ ++ γ − arity(g) +1

(LMCS, p. 141) III.18 Then s is a term iff the value of γ is always less than arity(f) except at the last symbol, where γ has the value arity(f). If s is a term then s = ft1 · · · tk where k = arity(f). The end of ti is the first symbol where γ is i.

(LMCS, p. 141) III.19 Example Let L = {f, g, c} with f unary g binary c a constant. Determine if s = ggcxfz is a term. If so find the subterms t1, t2 such that gt1t2 = ggcxfz. Here is the computation of γ: i 1 2 3 4 5 si g g c x f z γi −1 1 1 2

(LMCS, p. 142-143) III.20 Examples

• f the tree of a term:

(f is ternary, g binary)

fxgxyz + 1 (x y + ) ( ( y z)) +

+ y + y + z 1 x f x z g x y

(LMCS, p. 143) III.21 Looking at the tree of a term we see that it is built up in stages called subterms. Example Using infix notation, the subterms of ((x + y) · (y + z)) + 1 are x y z 1 x + y y + z (x + y) · (y + z) ((x + y) · (y + z)) + 1

+ x y z + y + 1

(LMCS, p. 143) III.22 Example Using prefix notation, the subterms of fxgxyz , where f is ternary and g is binary, are x y z gxy fxgxyz

f x z g x y

(LMCS, p. 143) III.23 Subterms The subterms of a term t are defined inductively:

• The only subterm of a variable

x is the variable x itself.

• The only subterm of a constant symbol

c is the symbol c itself.

• The subterms of

ft1 · · · tn are ft1 · · · tn itself and all the subterms of the ti, for 1 ≤ i ≤ n.

(LMCS, p. 145) III.24 Term Functions We interpret terms in an algebra as functions: terms t(x1, . . . , xn)

define

functions tA : An = ⇒ A Example Using the usual language for the natural numbers, consider the term t(x, y, z) given by (x · (y + 1)) + z The corresponding term function tN : N3 = ⇒ N maps the triple (1, 0, 2) to 3 as tN(1, 0, 2) = (1 · (0 + 1)) + 2 = 3

(LMCS, p. 145) III.25 Term functions tA for terms t(x1, . . . , xn) are the functions on the algebra A defined inductively by the following:

• If

t is xi then tA(a1, . . . , an) = ai.

• If t is

c ∈ C then tA(a1, . . . , an) = cA.

• If

t is ft1 · · · tk then tA = fA(tA

1 , . . . , tA k ).

(LMCS, p. 145) III.26 Evaluation Tables Example Let A be the two–element Boolean algebra

• n

{0, 1}, and let t(x, y, z) be the term x ∨ (y ∧ z′). Then the term function tA can be described by: x y z t 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

• r,

in more detail x y z z′ y ∧ z′

t

• x ∨ (y ∧ z′)

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

(LMCS, p. 149) III.27 Equations An L-equation is an expression of the form s ≈ t where s and t are L–terms. For example: x + y ≈ y + x. We think of this as saying for all x and y, x + y = y + x. = simply means “identical to”. ≈ expresses equality in formulas of our logics.

(LMCS, p. 149) III.28 The Basic Question: When does an equation follows from other equations? From x + y ≈ y + x we can conclude x + (y + z) ≈ (z + y) + x but not x + x ≈ x + y

(LMCS, p. 149) III.29 Two vantage points:

• (semantic) a notion of when an equation

is true in an algebra;

• (syntactic) axioms and rules of inference

for deriving equational consequences of equations. When you see the word “syntatic” think strings of symbols. Axioms are strings of symbols. Rules of Inference tell you how to take strings

• f symbols and make new strings of symbols.

(LMCS, p. 149) III.30 The semantic and syntatic notions of equational logic are equivalent! This equivalence is the soundness and completness of equational logic. Soundness means that one can only derive equations that correspond to valid arguments. Completeness means that one can derive all equations that correspond to valid arguments.

(LMCS, p. 149) III.31 Semantics Suppose L is a language of algebras.

• For A

an L–algebra and s ≈ t an L–equation we write

A |

= s ≈ t and say

A

satisfies s ≈ t or s ≈ t holds in A if sA = tA, that is, s and t define the same term function on A.

(LMCS, p. 149) III.32

• If

A

is an L–algebra and S is a set of L–equations, then we write

A |

= S and say

A

satisfies S

• r

S holds in A if A | = s ≈ t for every equation s ≈ t in S.

(LMCS, p. 149) III.33

• If

S is a set of L–equations and s ≈ t is an L–equation, then we write S | = s ≈ t and say s ≈ t is a consequence of S

• r

s ≈ t follows from S if

A |

= s ≈ t whenever

A |

= S. For satisfies we also use the word models, and we say A is a model of an equation or a set of equations.

(LMCS, p. 149) III.34 Laws of Binary Algebras A binary algebra A = (A, ·) is associative if it satisfies the associative law x · (y · z) ≈ (x · y) · z In such case it is also called a semigroup . It is commutative if it satisfies the commutative law x · y ≈ y · x And it is idempotent if it satisfies the idempotent law x · x ≈ x.

(LMCS, p. 150) III.35 The idempotent law is easy to check:

• ne looks down the main diagonal of the

table of the operation to see that x · x always has the value x.

a b c a b c a b c

(LMCS, p. 150) III.36 The commutative law is also easy to check: look at the table of the operation to see if it is symmetric about the main diagonal.

a b c a b c d d

(LMCS, p. 151) III.37 Example The binary algebra A given by · a b a a a b a b is idempotent, commutative, and associative:

A

| = x · x ≈ x

A

| = x · y ≈ y · x

A

| = x · (y · z) ≈ (x · y) · z. The first two properties follow by the preceding remarks and diagrams. To check the associative law we construct an evaluation table for the terms x · (y · z) and (x · y) · z to show the term functions are equal:

(LMCS, p. 151) III.38 From the binary algebra given by · a b a a a b a b we have the following table: x y z y · z

s

• x · (y · z)

x · y

t

• (x · y) · z

1. a a a a a a a 2. a a b a a a a 3. a b a a a a a 4. a b b b a a a 5. b a a a a a a 6. b a b a a a a 7. b b a a a b a 8. b b b b b b b Since the columns for x · (y · z) and (x · y) · z are the same, the associative law holds.

(LMCS, pp. 151-152) III.39 Example The binary algebra A given by · a b a b a b b b is not idempotent nor commutative nor associative. To see the failure of the associative law we use an evaluation table: x y z y · z

s

• x · (y · z)

x · y

t

• (x · y) · z

1. a a a b a b b 2. a a b a b b b 3. a b a b a a b 4. a b b b a a a 5. b a a b b b b 6. b a b a b b b 7. b b a b b b b 8. b b b b b b b Indeed, associativity fails precisely on lines 1 and 3.

(LMCS, pp. 151-152) III.40 However, this binary algebra · a b a b a b b b does satisfy an interesting equation, namely,

A |

= (x · y) · z ≈ (x · z) · y. We can see this by using an evaluation table: x y z x · y

s

• (x · y) · z

x · z

t

• (x · z) · y

a a a b b b b a a b b b a b a b a a b b b a b b a a a a b a a b b b b b a b b b b b b b a b b b b b b b b b b b

(LMCS, p. 153) III.41 Many interesting classes of algebras are defined by equations. The text gives examples of:

• Rings
• Boolean Algebras
• Semigroups
• Monoids
• Groups

You will NOT be required to memorize the equations defining these classes. They are for reference only.

(LMCS, p. 154) III.42 Rings The language LR is {+, ·, −, 0, 1}. R is the following set of equations in this language: R1. x + 0 ≈ x additive identity R2. x + (−x) ≈ additive inverse R3. x + y ≈ y + x + is commutative R4. x + (y + z) ≈ (x + y) + z + is associative R5. x · 1 ≈ x right mult. identity R6. 1 · x ≈ x left mult. identity R7. x · (y · z) ≈ (x · y) · z · is associative R8. x · (y + z) ≈ (x · y) + (x · z) left distributive R9. (x + y) · z ≈ (x · z) + (y · z) right distributive. All algebras R = (R, +, ·, −, 0, 1) that satisfy R are called rings. R is called a set of axioms or a set of defining equations for rings.

(LMCS, p. 154-155) III.43 Boolean Algebras We choose BA to be the following set of equations in the language LBA: B1. x ∨ y ≈ y ∨ x commutative B2. x ∧ y ≈ y ∧ x commutative B3. x ∨ (y ∨ z) ≈ (x ∨ y) ∨ z associative B4. x ∧ (y ∧ z) ≈ (x ∧ y) ∧ z associative B5. x ∧ (x ∨ y) ≈ x absorption B6. x ∨ (x ∧ y) ≈ x absorption B7. x ∧ (y ∨ z) ≈ (x ∧ y) ∨ (x ∧ z) distributive B8. x ∨ x′ ≈ 1 B9. x ∧ x′ ≈ B10. x ∨ 1 ≈ 1 B11. x ∧ 0 ≈ 0. All algebras B = (B, ∨, ∧,′ , 0, 1) that satisfy BA are called Boolean algebras. BA is called a set of axioms or a set of defining equations for Boolean algebra.

(LMCS, p. 156) III.44 Semigroups The language LSG = {·} consists of a single binary operation. SG has only one equation, the associative law: SG1 : (x · y) · z ≈ x · (y · z). Models of SG are called semigroups. SG axiomatizes or defines the class of semigroups.

(LMCS, p. 156-157) III.45 Monoids LM is {·, 1}, where · is binary and 1 is a constant symbol. M consists of: MO1 : x · 1 ≈ x MO2 : 1 · x ≈ x MO3 : (x · y) · z ≈ x · (y · z) . Any algebra A that satisfies M is called a monoid. M is a set of axioms or defining equations for monoids.

(LMCS, p. 157-158) III.46 Groups LG = {·, −1, 1}, where · is binary,

−1

is unary, and 1 is a constant symbol. G is the set of equations: G1 : x · 1 ≈ x G2 : x · x−1 ≈ 1 G3 : (x · y) · z ≈ x · (y · z). There is also an additive notation for groups, namely, the language is {+, −, 0} and is usually reserved for groups that are commutative: G1′ : x + 0 ≈ x G2′ : x + (−x) ≈ G3′ : (x + y) + z ≈ x + (y + z) G4′ : x + y ≈ y + x.

(LMCS, p. 158-159) III.47 Three Very Basic Properties of Equations (≈ behaves like an equivalence relation)

• A |

= s ≈ s

• A |

= s ≈ t implies

A |

= t ≈ s

• A |

= s1 ≈ s2 and

A |

= s2 ≈ s3 implies A | = s1 ≈ s3.

(LMCS, p. 158-159) III.48 Similar results hold for consequences. (Recall the definition of S | = s ≈ t from slide III.33, namely that any algebra that satisfies S will also satisfy s ≈ t.)

• S |

= s ≈ s

• S |

= s ≈ t implies S | = t ≈ s

• S |

= s1 ≈ s2 and S | = s2 ≈ s3 implies S | = s1 ≈ s3.

(LMCS, p. 161) III.49 Arguments that are Valid in

A

Given an algebra A, an argument s1 ≈ t1, · · · , sn ≈ tn

∴ s ≈ t,

is valid in A (or correct in A) provided either

• some equation

si ≈ ti does not hold in A,

• r
• s ≈ t

holds in A, i.e., if all the premisses hold in A, then so does the conclusion.

(LMCS, p. 163) III.50 Let A be the two–element Boolean algebra. To check the validity of the argument x ∨ y ≈ y ∨ x, x ∧ (y ∨ x) ≈ x

∴ x ∨ x′ ≈ x ∧ x′

we form an evaluation table: s1 t1 s2 t2 s t x y x ∨ y y ∨ x x ∧ (y ∨ x) x x ∨ x′ x ∧ x′ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Thus this is not a valid argument in A.

(LMCS, p. 163) III.51 Valid Equational Arguments An argument s1 ≈ t1, · · · , sn ≈ tn

∴ s ≈ t,

is valid (or correct) provided it is valid in every algebra A. Of course this becomes impossible to check using evaluation tables because there are an infinite number of algebras to examine.

(LMCS, p. 163) III.52 So how do we ever verify that an equational argument is valid? There are two ways. One is to use a proof system that we will soon study. The other is to study abstract algebra to learn special methods that can aid in a semantic analysis of validity.

(LMCS, p. 164) III.53 Refuting an Equational Argument An equational argument s1 ≈ t1, · · · , sn ≈ tn

∴ s ≈ t

is not valid iff one can find an algebra A such that

• all the premisses hold in A,
• but the conclusion does not hold.

Such an A is called a counterexample to the argument.

(LMCS, p. 164) III.54 One-Element Algebras We can never use a one–element algebra to refute an equational argument because in a

• ne–element algebra all equations are true

(there is only one value for the term functions to take). So to find a counterexample to an equational argument one must look to algebras with more than one element.

(LMCS, p. 164) III.55 Example Show that the argument x · y ≈ y · x

∴ x · x ≈ x

is not valid. We need to find a binary algebra that is commutative but not idempotent. The following two–element binary algebra does the job: · a b a a a b a a

(LMCS, p. 164-165) III.56 Example Show that the argument fffx ≈ fx

∴ ffx ≈ fx

is not valid. The following two–element monounary algebra provides a counterexample: f a b b a Such examples we usually discover by drawing pictures such as the following: f

a b

(LMCS, p. 168) III.57 What is Substitution? The word substitution encompasses two quite distinct kinds of substitution. One is for uniform substitution for

• variables. For example from

x + y ≈ y + x (*) we can deduce (x · y) + (z + z) ≈ (z + z) + (x · y) by substituting in (*) as follows: the term x · y for every occurrence of x the term z + z for every occurrence of y.

(LMCS, p. 168) III.58 The other use of substitution is to substitute

• ne term for another.

For example from x + y ≈ y + x we can also deduce (u · (x + y)) + (z + z) ≈ (u · (y + x)) + (z + z) Notice that we replaced the underlined term

• n the left with the underlined term on the

right.

(LMCS, p. 168) III.59 Since these two kinds of substitution are so different we will use the word substitution

• nly for the first kind, that is, for the uniform

substitution for variables; and call the second kind of substitution replacement.

(LMCS, p. 168) III.60 Substitution Given a term s(x1, . . . , xn) and terms t1, . . . , tn, the expression s(t1, . . . , tn) denotes the result of simultaneously substituting ti for xi in s. The notation we use for the substitution is

  

x1 ← t1 . . . xn ← tn

   .

(LMCS, p. 168) III.61 Example Let s(x, y) be x + y . Applying the substitution

• x ← x · z

y ← x + y

• to

s(x, y) gives s(x · z, x + y) which is (x · z) + (x + y)

(LMCS, p. 168-169) III.62 We can illustrate the substitution procedure with a picture.

+ x y y x z x + + s(x,y) = x+y s(x z , x+y) = (x z) + (x+y)

As you can see, substitution makes the tree grow downwards from the bottom. Substitution can only increase the tree size.

(LMCS, p. 169) III.63 Substitution Theorems

A |

= p(x1, . . . , xn) ≈ q(x1, . . . , xn) implies

A |

= p(t1, . . . , tn) ≈ q(t1, . . . , tn). S | = p(x1, . . . , xn) ≈ q(x1, . . . , xn) implies S | = p(t1, . . . , tn) ≈ q(t1, . . . , tn).

(LMCS, p. 171-172) III.64 Replacement Starting with a term p(· · · s · · · ) with an

• ccurrence of a subterm

s in it, we replace s with another term t and

• btain

p(· · · t · · · ). Example Replacing the second occurrence (from the left) of x + y in ((x + y

first

) · y) + (y + (x + y

second

)) with the term u · v gives the term ((x + y) · y) + (y + (u · v))

(LMCS, p. 171-172) III.65 We can visualize this by the use of trees:

+ y y + + + x y x y

+ ((x+y) ) y x ( y+( )) y +

y y + + + y x u v

((x+y) y) + ( y+( )) u v

Replacement has the effect of replacing one part of the tree. The replacement part can be larger or smaller than the original. Replacement can increase or decrease the size of the tree.

(LMCS, p. 173-174) III.66 Replacement Theorems

A |

= s ≈ t implies

A |

= p(· · · s · · · ) ≈ p(· · · t · · · ). S | = s ≈ t implies S | = p(· · · s · · · ) ≈ p(· · · t · · · ).

(LMCS, p. 175) III.67 The Syntactic Viewpoint Birkhoff’s Rules

Rule Name Example s ≈ s Reflexive x + y ≈ x + y s ≈ t t ≈ s Symmetric x ≈ x · x x · x ≈ x r ≈ s, s ≈ t r ≈ t Transitive x ≈ x · x, x · x ≈ 1 x ≈ 1 r( x) ≈ s( x) r( t ) ≈ s( t ) Substitution x · 1 ≈ x (x · x) · 1 ≈ x · x s ≈ t r(· · · s · · · ) ≈ r(· · · t · · · ) Replacement x · x ≈ x (x · x) + x ≈ x + x

(LMCS, p. 175) III.68 Derivation of Equations A derivation of an equation s ≈ t from S is a sequence s1 ≈ t1, . . . , sn ≈ tn

• f equations such that each equation is either

(1) a member of S, or (2) is the result of applying a rule of inference to previous members of the sequence, and the last equation is s ≈ t. We write S ⊢ s ≈ t if such a derivation exists.

(LMCS, p. 176) III.69 Example A derivation to witness fffx ≈ fy ⊢ fx ≈ fy is given by 1. fffx ≈ fy given 2. fffx ≈ fx subs 1 3. fx ≈ fffx symm 2 4. fx ≈ fy trans 1,3

(LMCS, p. 176) III.70 Example Show fgx ≈ x, ggx ≈ x ⊢ fx ≈ gx. 1. fgx ≈ x given 2. ggx ≈ x given 3. fggx ≈ gx subs 1 4. fggx ≈ fx repl 2 5. fx ≈ fggx symm 4 6. fx ≈ gx trans 3,5

(LMCS, p. 176-177) III.71 Example Show R ⊢ x · 0 ≈ 0.

1. x + 0 ≈ x given (R1) 2. 0 + 0 ≈ 0 subs 1 3. x · (0 + 0) ≈ x · 0 repl 2 4. x · (y + z) ≈ x · y + x · z given (R8) 5. x · (0 + 0) ≈ x · 0 + x · 0 subs 4 6. x · 0 + x · 0 ≈ x · (0 + 0) symm 5 7. x · 0 + x · 0 ≈ x · 0 trans 3,6 8. (x · 0 + x · 0) + (−(x · 0)) ≈ x · 0 + (−(x · 0)) repl 7 9. x + (−x) ≈ 0 given (R2)

• 10. x · 0 + (−(x · 0))

≈ 0 subs 9

• 11. (x · 0 + x · 0) + (−(x · 0)) ≈ 0

trans 8,10

• 12. x + (y + z)

≈ (x + y) + z given (R4)

• 13. x · 0 + (x · 0 + (−(x · 0))) ≈ (x · 0 + x · 0) + (−(x · 0)) subs 12
• 14. x · 0 + (x · 0 + (−(x · 0))) ≈ 0

trans 11,13

• 15. x · 0 + (−(x · 0))

≈ 0 subs 9

• 16. x · 0 + (x · 0 + (−(x · 0))) ≈ (x · 0) + 0

repl 15

• 17. x · 0 + 0

≈ x · 0 subs 1

• 18. x · 0 + (x · 0 + (−(x · 0))) ≈ x · 0

trans 16,17

• 19. x · 0

≈ x · 0 + (x · 0 + (−(x · 0))) symm 18

• 20. x · 0

≈ 0 trans 14,19

(LMCS, p. 183-185) III.72 Soundness and Completeness Birkhoff’s Rules are

• sound:

S ⊢ s ≈ t implies S | = s ≈ t

• complete:

S | = s ≈ t implies S ⊢ s ≈ t Propositional logic has many competing proof systems. In equational logic we prefer Birkhoff’s rules.

(LMCS, p. 183-185) III.73 Determining Validity In the propositional logic we have (often very slow) methods to check the validity of an argument. Equational logic has no such algorithm! This does not mean that we have not yet found an algorithm, but rather that it simply does not exist.

(LMCS, p. 185-186) III.74 Summary of Our Methods for Analyzing Arguments

• An equational argument

s1 ≈ t1, · · · , sn ≈ tn

∴ s ≈ t

is valid iff there is a derivation of the conclusion from the premisses, that is, iff s1 ≈ t1, · · · , sn ≈ tn ⊢ s ≈ t. Thus we look for a derivation to show an argument is valid.

(LMCS, p. 185-186) III.75

• An equational argument

s1 ≈ t1, · · · , sn ≈ tn

∴ s ≈ t

is invalid iff there is a counterexample A, that is, iff some

A

makes the premisses true and the conclusion false. Thus we can show an argument is not valid by finding a counterexample.

(LMCS, p. 190) III.76 Unification One of the most popular and powerful tools

• f automated theorem proving is an algorithm

to find most general unifiers. Let s(x1, . . . , xn) and s′(x1, . . . , xn) be two

• terms. A unifier of s and s′ is a substitution

  

x1 ← t1 . . . xn ← tn

  

such that s(t1, . . . , tn) = s′(t1, . . . , tn). After the substitution has been carried out, the two terms have become the same term. If a unifier can be found for s and t, we say they can be unified, or they are unifiable.

(LMCS, p. 190) III.77 Example Let s(x, y) = (x + y) · x t(x, y) = (y + x) · y. There are many unifiers of s and t, e.g.,

• x ← 1

y ← 1

• and
• x ← u · u

y ← u · u

• .

However the substitution

• x ← y

y ← y

• is a unifier that is more general than the

preceding two examples.

(LMCS, p. 191) III.78 Most General Unifiers If two terms s, t have a unifier µ such that every unifier of s, t is an instance of µ then we say µ is a most general unifier of s, t. Theorem (Robinson, 1965)

• If two terms are unifiable then they have a

most general unifier.

• The most general unifier of two terms is

(essentially) unique.

• There is an algorithm to determine if two

terms are unifiable, and if so the algorithm produces the most general unifier.

(LMCS, p. 192) III.79 Critical Subterms of s, t are the first subterms where s, t disagree. Example For the terms s = (x · y) + ((x + y) · x) t = (x · y) + ((y · (x + y)) · y) the critical subterms are s′ = x + y t′ = y · (x + y), x y x y x y x y y y x + + + + t = s =

(LMCS, p. 192-193) III.80 Critical Subterm Condition (CSC) The critical subterm condition (CSC) is satisfied by s and t if their critical subterms s′ and t′ consist of:

• a variable, say

x, and

• another term, say

r, that has no

• ccurrence of

x in it.

(LMCS, p. 192-193) III.81 Unification Algorithm: let µ be the identity substitution WHILE s = t find the critical subterms s′, t′ if the CSC fails, return NOT UNIFIABLE else with {s′, t′} = {x, r} apply (x ← r) to both s and t apply (x ← r) to µ ENDWHILE return (µ)

(LMCS, p. 193) III.82 A picture of a possible single step of the unification algorithm when the CSC holds:

s = t = r x ) ( x r r r

In this single step the substitution µ changes to (x ← r)µ.

(LMCS, p. 195) III.83 Example Apply the unification algorithm to x + (y · x) and (y · y) + z: s = t =

+ x y x + y y z x y y + y y z + y y y y y z y (y y) + y y y y y + y y y y y z y (y y) µ = x y y

(LMCS, p. 198-199) III.84 Composition of Substitutions Given two substitutions σ, σ′, the composition of the two is defined by (σ′σ)(t) = σ′(σ(t)) Example σ =

  

x ← x + y y ← x · y

  

and σ′ =

  

x ← y + x y ← x + x

  

leads to σ′σ =

  

x ← (y + x) + (x + x) y ← (y + x) · (x + x)

   .

(LMCS, p. 208) III.85 Term Rewrite Systems (TRS’s) A term rewrite (rule) is an expression s − → t, sometimes called a directed equation, where s and t are terms. A term rewrite system, abbreviated TRS, is a set R

• f term rewrite rules.

Example A simple system R

• f three term rewrite

rules used for monoids is R =

          

(x · y) · z − → x · (y · z) 1 · x − → x x · 1 − → x

          

.

(LMCS, p. 208) III.86 Elementary Rewrites An elementary rewrite obtained from s − → t is a rewrite of the form r(· · · σs · · · ) − → r(· · · σt · · · ), where σ is a substitution, r a term, and

• ne occurrence of

σs

• n the left has been

replaced by σt

• n the right.

Example 1 · (x · (y · z)) − → (x · (y · z)) is an elementary rewrite obtained from 1 · x − → x by using the substitution

• x ← x · (y · z)

(LMCS, p. 208-209) III.87 If R is a set of term rewrite rules then s − →R t means s − → t is an elementary rewrite of some term rewrite p − → q in R. We write s − →+

R t

if there is a finite sequence of elementary rewrites such that s = t1 − →R t2 − →R · · · − →R tn = t. The notation s − →∗

R t

means s = t

• r

s − →+

R t

holds.

(LMCS, p. 208-209) III.88 Example Using the monoid rules R we have ((x · 1) · (1 · y)) · (z · 1) − →+

R x · (y · z),

since ((x · 1) · (1 · y)) · (z · 1) − →R (x · (1 · y)) · (z · 1) (x · (1 · y)) · (z · 1) − →R (x · y) · (z · 1) (x · y) · (z · 1) − →R (x · y) · z (x · y) · z − →R x · (y · z)

(LMCS, p. 209) III.89 Terminating TRS’s

• A TRS

R is terminating if every sequence of elementary rewrites s1 − →R s2 − →R · · · eventually stops.

• A term

s is terminal if no elementary rewrite s − →R t is possible.

(LMCS, p. 209) III.90 Example The monoid example R is terminating. The terminal terms are either 1

• r of the

form x · (y · (z · · · w) · · · )) For example we have the terminating sequence ((x · y) · u) · v − →R (x · y) · (u · v) − →R x · (y · (u · v))

(LMCS, p. 209) III.91 Two Necessary Conditions for Terminating

• No rule of

R is of the form x − → t. For otherwise this could be applied to any term s, so no term would be terminal.

• If

s − → t is in R then the variables of t are also variables of s. For otherwise a substitution into s − → t would give an elementary rewrite of the form s − → t′ that has s as a subterm of t′, and this would permit an infinitely long sequence

• f rewrites.

(LMCS, p. 210) III.92 The Length of a Term |t| is the length of t (the number of symbols in the prefix form of t). |t|x is the x–length of t, the number of times the variable x

• ccurs in

t. Thus for t = (x · y) + (x · z) we have |t| = 7 |t|x = 2

(LMCS, p. 210) III.93 Theorem Let R be a term rewrite system with the property that for each s − → t ∈ R we have

• |s| > |t|,

and

• |s|x ≥ |t|x

for every variable x. Then R is a terminating TRS. (In this situation all of the elementary rewrites s − →R t are length-reducing.)

(LMCS, p. 211) III.94 Applications The following TRS’s are terminating:

• R = {x · x −

→ 0}

• R = {x · 1 −

→ x, 1 · x − → x}

• R = {x ∨ x′ −

→ x}

• R = {x ∨ (x ∧ y) −

→ x ∧ y}

• R = {x′′ ∨ x′ −

→ x ∨ x′}

• R = {fggx −

→ gfx, fgx − → fx}.

(LMCS, p. 211) III.95 Normal Form TRS’s A normal form TRS is a uniquely terminating TRS. For such a TRS the terminating form for any given term s is called the normal form of s, and written nR(s). A normal form TRS R is a normal form TRS for a set E of equations if E | = s ≈ t iff nR(s) = nR(t).

(LMCS, p. 211) III.96 Example R = {ffx − → x} is a normal form TRS. The normal form of fffx is fx. Example R =

    

(x · y) · z − → x · (y · z) x · 1 − → x 1 · x − → x

    

is a normal form TRS (for monoids). The normal form of ((x · 1) · (1 · y)) · z is x · (y · z).

(LMCS, p. 212-213) III.97 A Normal Form TRS for Groups (Knuth and Bendix, 1967) E R 1 · x ≈ x 1−1 − → 1 x−1 · x ≈ 1 x · 1 − → x (x · y) · z ≈ x · (y · z) 1 · x − → x (x−1)−1 − → x x−1 · x − → 1 x · x−1 − → 1 x−1 · (x · y) − → y x · (x−1 · y) − → y (x · y)−1 − → y−1 · x−1 (x · y) · z − → x · (y · z) In this case it is not immediate that the system R is terminating, much less a normal form TRS.

(LMCS, p. 213) III.98 Converting R into an equational theory Let E(R) = {s ≈ t : s − → t ∈ R}. Proposition If R is a normal form TRS, then it is a normal form TRS for E(R). Thus, for example, once we show that R = {(x · y) · z − → x · (y · z)} is a normal form TRS, then it follows that it is a normal form TRS for semigroups.

(LMCS, p. 215) III.99 Critical Pairs Given s1 − → t1 s2 − → t2. Rename the variables in one of them (if necessary) so that s1 and s2 have no variables in common. Choose an occurrence of a nonvariable subterm s′

1

• f

s1 such that s′

1, s2

are unifiable. Find the most general unifier µ

• f

s′

1

and s2. Two rewrites of µs1 give a critical pair.

(LMCS, p. 216) III.100 t 2

1

s s = s = t 1

1 2

s =

2

µ t 1 µ µ 1 s s =

1

µ t 2 µ

Given: Two Rewrite Rules with disjoint variables

2 1

s s µ Then apply it to both rules. Find , the most general unifier of and .

(LMCS, p. 216) III.101 Creating the Critical Pair

t 1 µ µ 2 t = µ 1 s µ s

2

THIS IS A CRITICAL PAIR

Use the second rule

Use the first rule

(LMCS, p. 216) III.102 Example: Find the critical pair for y + (f(x) + y) − → f(x + y) x + g(y) − → g(x + y) Changing variables we have y + (f(x) + y) − → f(x + y) u + g(v) − → g(u + v) A most general unifier: µ =

  

u ← f(x) y ← g(v)

   .

Apply this to the original rewrites: g(v) + (f(x) + g(v)) − → f(x + g(v)) f(x) + g(v) − → g(f(x) + v). Thus the critical pair is: (f(x + g(v)) and g(v) + g(f(x) + v)).

(LMCS, p. 223-224) III.103 A TRS R is confluent on critical pairs if, given any critical pair (s, t), there is a term r such that s − →∗

R

r t − →∗

R

r. Critical Pairs Lemma (Knuth and Bendix) Suppose R is a terminating TRS. Then R is a normal form TRS iff R is confluent on critical pairs.

(LMCS, p. 223-224) III.104 Corollary Let R be a term rewrite system with the property that for each s − → t ∈ R we have

• |s| > |t|,
• |s|x ≥ |t|x

for every variable x, and

• R

is confluent on critical pairs. Then R is a normal form TRS.

(LMCS, p. 224) III.105 Example Consider R = {fgfx − → gx}, a terminating TRS. fg fx − → gx fgfu − → gu has µ = (x ← gfu). Applying µ: fg fgfu − → ggfu fgfu − → gu The critical pair is: (ggfu, fggu). We do not have confluence for this critical pair, so the system R is not a normal form TRS.