Propositions 2/57 A proposition is a statement that is true or - - PowerPoint PPT Presentation

SLIDE 1

/ department of mathematics and computer science

Propositional Logic

Lectures 1 and 2 (Chapters 2–7)

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Propositions

A proposition is a statement that is true or false.

Examples:

◮ grass is green ◮ 5 > 3 ◮ 5 < 3 ◮ grass is green and roses are blue ◮ if x > 1, then x2 = x

Non-examples:

◮ What time is it? ◮ Don’t look back! 3/57 / department of mathematics and computer science

Abstract propositions: vocabulary

We want to study logic without being distracted by the concrete contents of propositions!

Proposition variables:

a, b, c, . . .

Connectives:

∧ ’and’ ∨ ’or’ ¬ ’not’ ⇒ ’if . . . then . . . ’ ⇔ ’if and only if’

Examples:

◮ grass is green

a

◮ 5 > 3

a

◮ 5 < 3

a

◮ grass is green and

roses are blue a ∧ b

◮ if x > 1, then x2 = x

a ⇒ ¬b

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Abstract propositions: grammar

Inductive definition of abstract propositions:

• 1. (BASIS) every proposition variable is an abstract proposition;
• 2. (STEP)

2.1 if P is an abstract proposition, then so is (¬P); 2.2 if P and Q are abstract propositions, then so are (P ∧ Q), (P ∨ Q), (P ⇒ Q), and (P ⇔ Q).

(¬ is unary; ∧, ∨, ⇒, and ⇔ are binary)

Examples:

a b (¬a) ((¬a) ∧ b) (((¬a) ∧ b) ∨ b)

Non-examples:

a∧ ⇒⇒a a¬b

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Abstract propositions: grammar

Inductive definition of abstract propositions:

• 1. (BASIS) every proposition variable is an abstract proposition;
• 2. (STEP)

2.1 if P is an abstract proposition, then so is (¬P); 2.2 if P and Q are abstract propositions, then so are (P ∧ Q), (P ∨ Q), (P ⇒ Q), and (P ⇔ Q).

(¬ is unary; ∧, ∨, ⇒, and ⇔ are binary) Show that (((¬a) ∧ b) ∨ b) is indeed an abstract proposition:

1 a 2.1 (¬a) 1

b

2.2 ((¬a) ∧ b) 1

b

2.2

(((¬a) ∧ b) ∨ b)

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Omitting parentheses:

We want to omit as many parentheses from abstract propositions as possible, but without causing ambiguity.

• 1. Outermost parentheses can always be omitted
• 2. We agree on the following priority schema:

¬ ∧ ∨ ⇒ ⇔ NB: since ¬ has highest priority, parentheses around a negation may always be omitted. We may, e.g., also omit the parentheses from ¬(¬a).

Examples:

((¬a) ∨ (¬b))

1

• (¬a) ∨ (¬b)

2

• ¬a ∨ ¬b

((¬a) ∧ b)

1

• (¬a) ∧ b

2

• ¬a ∧ b

(¬(a ∧ b))

1

• ¬(a ∧ b)

?

• ?
• ¬a ∧ b

¬a ∧ b NO!

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Negation

A negation ¬P is true if P is false; false if P is true.

Truth table for ¬:

P ¬P 1 1 0 = false 1 = true

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Conjunction

A conjunction P ∧ Q is true if P is true and Q is true; false in all other cases.

Truth table for ∧:

P Q P ∧ Q 1 1 1 1 1 We first list all possible combinations of assignments to P and Q.

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Disjunction

A disjunction P ∨ Q is true if P is true, or Q is true, or both; false otherwise.

Truth table for ∨:

P Q P ∨ Q 1 1 1 1 1 1 1

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Inclusive vs. Exclusive Or

Examples:

◮ Inclusive:

Can you show me a passport or drivers license?

◮ Exclusive:

Do you want peanut butter or jam on your sandwich?

Exercise:

Give abstract proposition that ‘behaves’ as exclusive or of a and b (notation: a ⊕ b).

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Implication

An implication P ⇒ Q is true if whenever P is true, then also Q is true; false otherwise.

Truth table for ⇒:

P Q P ⇒ Q 1 1 1 1 1 1 1

When is P ⇒ Q clearly not true? Now consider n > 2 ⇒ n + 1 > 2. Clearly, this implication is true for every n. n = 1: both n > 2 and n + 1 > 2 false. n = 2: n > 2 false, n + 1 > 2 true. n = 3: both n > 2 and n + 1 > 2 true.

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Biimplication

A biimplication P ⇔ Q is true if P and Q have the same truth value; false otherwise.

Truth table for ⇔:

P Q P ⇔ Q 1 1 1 1 1 1

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Computing truth table

Example:

Computing the truth table of ¬(a ⇒ b): a b a ⇒ b ¬(a ⇒ b) 1 1 1 1 1 1 1 1

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Tautology

An abstract proposition is a tautology if its column in a truth table exclusively consists of 1s.

Examples:

◮ a ∨ ¬a ◮ a ⇒ a ◮ a ⇒ (b ⇒ a) ◮ . . .

a b b ⇒ a a ⇒ (b ⇒ a) 1 1 1 1 1 1 1 1 1 1 1

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Contradiction

An abstract proposition is a contradiction if its column in a truth table exclusively consists of 0s.

Examples:

◮ a ∧ ¬a ◮ (a ⇒ b) ∧ (a ∧ ¬b) ◮ . . .

a b a ⇒ b ¬b a ∧ ¬b (a ⇒ b) ∧ (a ∧ ¬b) 1 1 1 1 1 1 1 1 1 1

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Contingency

An abstract proposition is contingent if it is not a tautology, nor a contradiction.

Examples:

• 1. a
• 2. a ⇒ ¬a
• 3. a ∧ b
• 4. a ∨ b
• 5. ¬a ⇒ (b ∧ c)

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Equivalence

Example:

Combine truth tables for ¬(a ⇒ b) and ¬(¬a ∨ b): a b a ⇒ b ¬(a ⇒ b) ¬a ¬a ∨ b ¬(¬a ∨ b) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Note: the columns for ¬(a ⇒ b) and ¬(¬a ∨ b) are identical. Abstract propositions with identical columns in a combined truth table are said to be equivalent.

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Equivalence

All tautologies are equivalent. We introduce an extra symbol True to denote an arbitrary tautology. All contradictions are equivalent. We introduce an extra symbol False to denote an arbitrary contradiction. Not all contingencies are equivalent.

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Abstract propositions: grammar (extended)

From now on, we shall consider True and False officially as part of the vocabulary of abstract propositions.

Inductive definition of abstract propositions:

• 1. (BASIS) True and False are abstract propositions, and every

proposition variable is an abstract proposition;

• 2. (STEP)

2.1 if P is an abstract proposition, then so is (¬P); 2.2 if P and Q are abstract propositions, then so are (P ∧ Q), (P ∨ Q), (P ⇒ Q), and (P ⇔ Q).

(¬ is unary; ∧, ∨, ⇒, and ⇔ are binary)

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Notation

If P is equivalent to Q, then we write P

val

= = Q. Note:

val

= = is not part of the vocabulary of the language of abstract propositions; it is a meta-symbol. So: a ⇒ b

• abstr. prop.

val

= = ¬a ∨ b

• abstr. prop.
• meta-formula

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Proving equivalences

Suppose we want to prove that ¬(Q ⇒ R)

val

= = ¬R ∧ Q. How did we proceed (until now)? We would prove the equivalence using truth tables: Q R Q ⇒ R ¬(Q ⇒ R) ¬R ¬R ∧ Q 1 1 1 1 1 1 1 1 1 1 1 The columns of ¬(Q ⇒ R) and ¬R ∧ Q are identical, so it holds that ¬(Q ⇒ R)

val

= = ¬R ∧ Q.

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Commutativity, Associativity

Commutativity: P ∧ Q

val

= = Q ∧ P P ∨ Q

val

= = Q ∨ P P ⇔ Q val = = Q ⇔ P Associativity: (P ∧ Q) ∧ R

val

= = P ∧ (Q ∧ R) (P ∨ Q) ∨ R

val

= = P ∨ (Q ∨ R) (P ⇔ Q) ⇔ R val = = P ⇔ (Q ⇔ R)

NB: P ⇒ Q

val

= = = Q ⇒ P P Q P ⇒ Q Q ⇒ P 0 1 1 NB: P ⇒ (Q ⇒ R)

val

= = = (P ⇒ Q) ⇒ R P Q R P ⇒ (Q ⇒ R) (P ⇒ Q) ⇒ R 0 1 0 1

NB: In view of Associativity, we shall write P ∧ Q ∧ R to denote both (P ∧ Q) ∧ R and P ∧ (Q ∧ R).

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Be Careful!

Are the abstract propositions (a ∧ b) ∨ c and a ∧ (b ∨ c) equivalent? NO! (see p. 19 in your book) This shows that the (remaining) parentheses in these abstract propositions are important. In fact, they make the difference!

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Idempotence, Double negation

Idempotence: P ∧ P val = = P P ∨ P val = = P NB: P ⇒ P

val

= = = P P ⇔ P

val

= = = P Double Negation: ¬¬P val = = P ‘It is not that I don’t like spinach’ (NB: in propositional logic the intended nuance cannot be captured.)

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True and False

Inversion: ¬True

val

= = False ¬False val = = True Negation: ¬P val = = P ⇒ False Contradiction: P ∧ ¬P val = = False Excluded Middle: P ∨ ¬P val = = True True/False-elimination: P ∧ True

val

= = P P ∧ False val = = False P ∨ True

val

= = True P ∨ False val = = P

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Distributivity and De Morgan

Distributivity: P ∧ (Q ∨ R) val = = (P ∧ Q) ∨ (P ∧ R) P ∨ (Q ∧ R) val = = (P ∨ Q) ∧ (P ∨ R) De Morgan: ¬(P ∧ Q) val = = ¬P ∨ ¬Q ¬(P ∨ Q) val = = ¬P ∧ ¬Q

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Implication and Contraposition

Implication: P ⇒ Q val = = ¬P ∨ Q ¬P ⇒ Q

val

= = { Implication } ¬¬P ∨ Q

val

= = { Double Negation } P ∨ Q Contraposition: P ⇒ Q val = = ¬Q ⇒ ¬P NB: P ⇒ Q

val

= = = ¬P ⇒ ¬Q (NB: ‘if . . . , then . . . ’ in natural language

• ften means ‘if and only if’.)

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Bi-implication and self-equivalence

Bi-implication: P ⇔ Q val = = (P ⇒ Q) ∧ (Q ⇒ P) Self-equivalence: P ⇔ P val = = True

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Table of standard equivalences

372 TABLES FOR PART I

Equivalences for connectives

Commutativity: Associativity: P ∧ Q

val

= = Q ∧ P, (P ∧ Q) ∧ R

val

= = P ∧ (Q ∧ R), P ∨ Q

val

= = Q ∨ P, (P ∨ Q) ∨ R

val

= = P ∨ (Q ∨ R), P ⇔ Q

val

= = Q ⇔ P (P ⇔ Q) ⇔ R

val

= = P ⇔ (Q ⇔ R) Idempotence: Double Negation: P ∧ P

val

= = P, ¬¬P

val

= = P P ∨ P

val

= = P Inversion: True/False-elimination: ¬True

val

= = False, P ∧ True

val

= = P, ¬False

val

= = True P ∧ False

val

= = False, P ∨ True

val

= = True, P ∨ False

val

= = P Negation: Contradiction: ¬P

val

= = P ⇒ False P ∧ ¬P

val

= = False Excluded Middle: P ∨ ¬P

val

= = True Distributivity: De Morgan: P ∧ (Q ∨ R)

val

= = (P ∧ Q) ∨ (P ∧ R), ¬(P ∧ Q)

val

= = ¬P ∨ ¬Q, P ∨ (Q ∧ R)

val

= = (P ∨ Q) ∧ (P ∨ R) ¬(P ∨ Q)

val

= = ¬P ∧ ¬Q Implication: Contraposition: P ⇒ Q

val

= = ¬P ∨ Q P ⇒ Q

val

= = ¬Q ⇒ ¬P Bi-implication: Self-equivalence: P ⇔ Q

val

= = (P ⇒ Q) ∧ (Q ⇒ P) P ⇔ P

val

= = True

For the collection of all standard equivalences, see page 372 of the book! You will have to know them by heart (including their names!). Start memorising them today!

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Calculation

Recall the following calculation: ¬P ⇒ Q

val

= = { Implication } ¬¬P ∨ Q

val

= = { Double Negation } P ∨ Q Can we conclude ¬P ⇒ Q

val

= = P ∨ Q ?

• 1. What about applying two standard equivalences in a row?

Does it preserve equivalence?

• 2. First step: not a literal application of Implication.

Can we do substitutions?

• 3. Second step: literal application of Double Negation.

Is it safe to apply standard equivalences in a larger context?

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val

= = is a decent equivalence

Lemma 6.1.1

• 1. (Reflexivity:) P

val

= = P

• 2. (Symmetry:) If P

val

= = Q, then Q

val

= = P

• 3. (Transitivity:) If P

val

= = Q and Q

val

= = R, then P

val

= = R

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Substitution

Substitution is the replacement of all occurrences of a ‘letter’ by a formula.

Examples:

• 1. If we substitute Q ∧ P for P in the valid equivalence

P ⇒ Q

val

= = ¬P ∨ Q , then we get the valid equivalence (Q ∧ P) ⇒ Q

val

= = ¬(Q ∧ P) ∨ Q .

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Substitution

Substitution is the replacement of all occurrences of a ‘letter’ by a formula.

Examples:

• 2. If we substitute ¬R for Q in the valid equivalence

(Q ∧ P) ⇒ Q

val

= = ¬(Q ∧ P) ∨ Q , then we get the valid equivalence (¬R ∧ P) ⇒ ¬R

val

= = ¬(¬R ∧ P) ∨ ¬R .

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Substitution

Substitution is the replacement of all occurrences of a ‘letter’ by a formula.

Examples:

• 3. If we (simultaneously) substitute Q ∧ P for P and ¬R for Q in the

valid equivalence P ⇒ Q

val

= = ¬P ∨ Q , then we get the valid equivalence (Q ∧ P) ⇒ ¬R

val

= = ¬(Q ∧ P) ∨ ¬R .

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Substitution Rule

SUBSTITUTION PRESERVES EQUIVALENCE

Important remarks:

• 1. Substitution operates on entire equivalences
• 2. If you substitute for some letter P in an equivalence, then you

have to replace all occurrences of P in that equivalence!

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Leibniz’s rule

Leibniz’s rule is about the replacement of a subformula by an equivalent subformula.

Example:

From the valid equivalence P ⇒ Q

val

= = ¬P ∨ Q we can make new valid equivalences by replacing P ⇒ Q in some complex formula by ¬P ∨ Q, for instance: (¬P ∧ (P ⇒ Q)) ∨ R

val

= = (¬P ∧ (¬P ∨ Q)) ∨ R

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Leibniz’s rule

Leibniz’s rule is about the replacement of a subformula by an equivalent subformula.

Schematically:

P

val

= = Q · · · P · · ·

val

= = · · · Q · · ·

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Proving tautologies—method 1

To prove with a calculation that P is a tautology: Give calculation that shows P

val

= = True.

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Proving tautologies—method 1 (example)

Prove with a calculation that ¬(P ∧ ¬P) is a tautology. We have the following calculation: ¬(P ∧ ¬P)

val

= = { De Morgan } ¬P ∨ ¬¬P

val

= = { Double Negation } P ∨ ¬P

val

= = { Excluded Middle } True So ¬(P ∧ ¬P) is a tautology.

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Proving tautologies—method 2

Lemma 6.1.3

If P

val

= = Q, then P ⇔ Q is a tautology, and vice versa. To prove with a calculation that P is a tautology: Give a calculation that shows P

val

= = Q.

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Proving tautologies—method 2 (example)

Prove with a calculation that ¬(Q ⇒ R) ⇔ (¬R ∧ Q) is a tautology.

First, we establish, with a calculation, that ¬(Q ⇒ R)

val

= = (¬R ∧ Q): ¬(Q ⇒ R)

val

= = { Implication } ¬(¬Q ∨ R)

val

= = { De Morgan } ¬¬Q ∧ ¬R

val

= = { Double negation } ¬R ∧ Q Explanation: Substituting Q for P and R for Q in P ⇒ Q

val

= = ¬P ∨ Q (Implication) we get, by the substitution rule: Q ⇒ R

val

= = ¬Q ∨ R. (The application of this equivalence in the calculation involves an application

• f Leibniz.)

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Proving tautologies—method 2 (example)

Prove with a calculation that ¬(Q ⇒ R) ⇔ (¬R ∧ Q) is a tautology.

First, we establish, with a calculation, that ¬(Q ⇒ R)

val

= = (¬R ∧ Q): ¬(Q ⇒ R)

val

= = { Implication } ¬(¬Q ∨ R)

val

= = { De Morgan } ¬¬Q ∧ ¬R

val

= = { Double negation } ¬R ∧ Q Explanation: Substituting ¬Q for P and R for Q in ¬(P ∨ Q)

val

= = ¬P ∧ ¬Q (De Morgan) we get, by the substitution rule: ¬(¬Q ∨ R)

val

= = ¬¬Q ∧ ¬R.

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Proving tautologies—method 2 (example)

Prove with a calculation that ¬(Q ⇒ R) ⇔ (¬R ∧ Q) is a tautology.

First, we establish, with a calculation, that ¬(Q ⇒ R)

val

= = (¬R ∧ Q): ¬(Q ⇒ R)

val

= = { Implication } ¬(¬Q ∨ R)

val

= = { De Morgan } ¬¬Q ∧ ¬R

val

= = { Double negation } ¬R ∧ Q Explanation: Substituting Q for P in ¬¬P

val

= = P (Double negation) we get, by the substitution rule: ¬¬Q

val

= = Q. (The application of this equivalence in the calculation involves an application

• f Leibniz, and is followed by an

application of Commutativity.)

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Proving tautologies—method 2 (example)

Prove with a calculation that ¬(Q ⇒ R) ⇔ (¬R ∧ Q) is a tautology.

First, we establish, with a calculation, that ¬(Q ⇒ R)

val

= = (¬R ∧ Q): ¬(Q ⇒ R)

val

= = { Implication } ¬(¬Q ∨ R)

val

= = { De Morgan } ¬¬Q ∧ ¬R

val

= = { Double negation } ¬R ∧ Q From ¬(Q ⇒ R)

val

= = ¬R ∧ Q it follows (by Lemma 6.1.3) that ¬(Q ⇒ R) ⇔ (¬R ∧ Q) is a tautology.

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Logical Consequence

Recall: P

val

= = Q means (a) whenever P is 1, then also Q is 1 (b) whenever Q is 1, then also P is 1 Define: P |

val

= = Q means

• (a) whenever P is 1, then also Q is 1

Pronounce P |

val

= = Q as “P is stronger than Q.” P Q 1 1 1/0 P |

val

= = Q: 1s are carried over from P to Q.

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Logical Consequence (example 1)

¬P

• |

val

= = ?

val

= = | ?

• P ⇒ Q

P Q ¬P P ⇒ Q 1 1 1 1 1 1 1 1 1 extra true So ¬P is stronger than P ⇒ Q (i.e., ¬P |

val

= = P ⇒ Q).

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Logical Consequence (example 2)

P ⇒ Q

• |

val

= = ?

val

= = | ?

• P ∨ Q

P Q P ⇒ Q P ∨ Q 1 1 1 1 1 1 1 1 1 1 So P ⇒ Q and P ∨ Q are incomparable.

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Standard weakenings

∧-∨-weakening: P ∧ Q |

val

= = P P |

val

= = P ∨ Q Also: P ∧ Q |

val

= = Q and Q |

val

= = P ∨ Q . P Q P ∧ Q P P ∨ Q 1 1 1 1 1 1 1 1 1 1 Extremes: False |

val

= = P P |

val

= = True False is strongest of all True is weakest of all

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Basic properties of |

val

= = ,

val

= = |

Lemma 7.3.1

(1a) P |

val

= = P. (2) If P |

val

= = Q, then Q

val

= = | P, and vice versa. (3) If P |

val

= = Q and Q |

val

= = R, then P |

val

= = R.

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Basic properties of |

val

= = ,

val

= = |

Lemma 7.3.2

P

val

= = Q if, and only if, P |

val

= = Q and P

val

= = | Q.

So, if you need to prove P |

val

= = Q or P

val

= = | Q by a calculation, then it is enough to prove P

val

= = Q. But P |

val

= = Q (or P

val

= = | Q) alone is not enough to conclude P

val

= = Q!

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Basic properties of |

val

= = ,

val

= = |

Lemma 7.3.4

P |

val

= = Q if, and only if, P ⇒ Q is a tautology.

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Substitution Rule for |

val

= = ,

val

= = |

The Substitution Rule also works for |

val

= = and

val

= = | : SUBSTITUTION PRESERVES WEAKENING/STRENGHTENING

Example

We have the following valid weakening: P ∧ Q |

val

= = P ∨ R and hence, according to the Substitution Rule, if we substitute (Q ⇒ R) for P and (P ∨ Q) for Q , we get another valid weakening: (Q ⇒ R) ∧ (P ∨ Q) |

val

= = (Q ⇒ R) ∨ R .

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Leibniz for |

val

= = ,

val

= = | ?

Recall Leibniz’s Rule for making new equivalences: P

val

= = Q · · · P · · ·

val

= = · · · Q · · · Can we replace

val

= = by |

val

= = in this rule?

Examples

Note that, by ∧-∨-weakening, P ∧ Q |

val

= = P ∨ Q. Now consider:

• 1. ¬(P ∧ Q) |

val

= = ¬(P ∨ Q);

• 2. R ⇒ (P ∧ Q) |

val

= = R ⇒ (P ∨ Q); and

• 3. (P ∧ Q) ⇒ R |

val

= = (P ∨ Q) ⇒ R. Conclusion: replacing

val

= = by |

val

= = does not yield a valid rule!

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Monotonicity

We do have the following weaker variant of Leibniz’s Rule: Monotonicity: (1) If P |

val

= = Q, then P ∧ R |

val

= = Q ∧ R (2) If P |

val

= = Q, then P ∨ R |

val

= = Q ∨ R

Example:

Since P |

val

= = P ∨ Q by ∧-∨-weakening, we have: P ∧ R |

val

= = { ∧-∨-weakening + Monotonicity } (P ∨ Q) ∧ R .

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Example

Prove with a calculation that ¬(P ⇒ Q) ⇒ (¬Q ∧ (P ∨ R)) is a tautology.

First, we establish that ¬(P ⇒ Q) |

val

= = ¬Q ∧ (P ∨ R): ¬(P ⇒ Q)

val

= = { Implication } ¬(¬P ∨ Q)

val

= = { De Morgan } ¬¬P ∧ ¬Q

val

= = { Double Negation } P ∧ ¬Q |

val

= = { ∧-∨-weakening + Monotonicity } ¬Q ∧ (P ∨ R) So, by Lemma 7.3.4, the formula ¬(P ⇒ Q) ⇒ (¬Q ∧ (P ∨ R)) is a tautology.

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Formal System for Calculation

We now have a precisely defined formal system for calculating with abstract propositions:

◮ standard equivalences and standard weakenings; ◮ inference rules (viz. reflexivity, symmetry, transitivity,

substitution, Leibniz for equality, Monotonicity for weakening) It gives a method to prove in a structured manner that

◮ two abstract propositions are equivalent, or one is

stronger/weaker than the other;

◮ an abstract proposition is a tautology.