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Propositional Logic: Semantics
Alice Gao
Lecture 4, September 19, 2017
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Announcements
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Propositional Logic: Semantics Alice Gao Lecture 4, September 19, - - PowerPoint PPT Presentation
Propositional Logic: Semantics Alice Gao Lecture 4, September 19, - - PowerPoint PPT Presentation
Propositional Logic: Semantics Alice Gao Lecture 4, September 19, 2017 Semantics 1/56 Announcements Semantics 2/56 The roadmap of propositional logic Semantics 3/56 FCC spectrum auction an application of propositional spectrums to
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The roadmap of propositional logic
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FCC spectrum auction β an application of propositional logic
To repurpose radio spectrums 2 auctions:
- one to buy back spectrums from broadcasters
- the other to sell spectrums to telecoms
A computational problem in the buy back auction: If I pay these broadcasters to go ofg air, could I repackage the spectrums and sell to telecoms? Could I lower your price and still manage to get useful spectrums to sell to telecoms? The problem comes down to, how many satisfjability problems can I solve in a very short amount of time? (determine that a formula is satisfjable or determine that it is unsatisfjable.) Talk by Kevin Leyton-Brown https://www.youtube.com/watch?v=u1-jJOivP70
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Learning goals
By the end of this lecture, you should be able to
- Evaluate the truth value of a formula
- Defjne a (truth) valuation.
- Determine the truth value of a formula by using truth tables.
- Determine the truth value of a formula by using valuation trees.
- Determine and prove whether a formula has a particular property
- Defjne tautology, contradiction, and satisfjable formula.
- Compare and contrast the three properties (tautology, contradiction,
and satisfjable formula).
- Prove whether a formula is a tautology, a contradiction, or satisfjable,
using a truth table and/or a valuation tree.
- Describe strategies to prove whether a formula is a tautology, a
contradiction or a satisfjable formula.
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The meaning of well-formed formulas
To interpret a formula, we have to give meanings to the propositional variables and the connectives. A propositional variable has no intrinsic meaning; it gets a meaning via a valuation. A (truth) valuation is a function π’ βΆ π¬ β¦ {F, T} from the set of all proposition variables π¬ to {F, T}. It assigns true/false to every propositional variable. Two notations: π’(π) and ππ’ both denote the truth value of π under the truth valuation π’.
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Truth tables for connectives
The unary connective Β¬: π½ (Β¬π½) T F F T The binary connectives β§, β¨, β, and β: π½ πΎ (π½ β§ πΎ) (π½ β¨ πΎ) (π½ β πΎ) (π½ β πΎ) F F F F T T F T F T T F T F F T F F T T T T T T
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Truth value of a formula
Fix a truth valuation π’. Every formula π½ has a value under π’, denoted π½π’, determined as follows.
- 1. ππ’ = π’(π).
- 2. (Β¬π½)π’ = { T
if π½π’ = F F if π½π’ = T
- 3. (π½ β§ πΎ)π’ = { T
if π½π’ = πΎπ’ = T F
- therwise
- 4. (π½ β¨ πΎ)π’ = { T
if π½π’ = T or πΎπ’ = T F
- therwise
- 5. (π½ β πΎ)π’ = { T
if π½π’ = F or πΎπ’ = T F
- therwise
- 6. (π½ β πΎ)π’ = { T
if π½π’ = πΎπ’ F
- therwise
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Evaluating a formula using a truth table
Example. The truth table of ((π β¨ π) β (π β§ π )). π π π (π β¨ π) (π β§ π ) ((π β¨ π) β (π β§ π )) F F F F F T F F T F F T F T F T F F F T T T T T T F F T F F T F T T F F T T F T F F T T T T T T
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Evaluating a formula using a truth table
Build the truth table of ((π β (Β¬π)) β (π β¨ (Β¬π))).
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Understanding the disjunction and the biconditional
π½ πΎ (π½ β¨ πΎ) Exclusive OR Biconditional F F F F T F T T T F T F T T F T T T F T
- What is the difgerence between an inclusive OR (the disjunction) and
an exclusive OR?
- What is the relationship between the exclusive OR and the
biconditional?
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Understanding the conditional β
Assume that proposition π defjned below is true. π: If Alice is rich, she will pay your tuition. If Alice is rich, will she pay your tuition?
- a. Yes
- b. No
- c. Maybe
If Alice is not rich, will she pay your tuition?
- a. Yes
- b. No
- c. Maybe
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Understanding the conditional β
Alice is rich Alice will pay your tuition. If Alice is rich, she will pay your tuition. F F T F T T T F F T T T
- Suppose that the implication is a promise that I made. How can you
show that I broke my promise?
- If the premise is false, is the statement true or false? (Will the
statement ever be contradicted?)
- When the conclusion is true, is the statement true or false?
- When the premise is true, how does the truth value of the statement
compare to the truth value of the conclusion?
- Convert π β π into a logically equivalent formula which only uses the
connectives β§, β¨ and Β¬. Does this alternative formula make sense?
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Another example of structural induction
Theorem: Fix a truth valuation π’. Every formula π½ has a value π½π’ in {F, T}. Proof: The property for π(π½) is βπ½ has a value π½π’ in {F, T}β.
- 1. If π½ is a propositional variable, then π’ assigns it a value of T or F (by
the defjnition of a truth valuation).
- 2. If π½ has a value in {F, T}, then (Β¬π½) also does, as shown by the truth
table of (Β¬π½).
- 3. If π½ and πΎ each has a value in {F, T}, then (π½ β πΎ) also does for every
binary connective β, as shown by the corresponding truth tables. By the principle of structural induction, every formula has a value. By the unique readability of formulas, we have proved that a formula has
- nly one truth value under any truth valuation π’. QED
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Tautology, Contradiction, Satisfjable
A formula π½ is a tautology if and only if for every truth valuation π’, π½π’ = T. A formula π½ is a contradiction if and only if for every truth valuation π’, π½π’ = F. A formula π½ is satisfjable if and only if there exists a truth valuation π’ such that π½π’ = T.
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Relationships among the properties
Divide the set of all formulas into 3 mutually exclusive and exhaustive sets. We know two things about these sets:
- A formula is in set 1 if and only if the formula is true in every row of
the formulaβs truth table.
- A formula is in set 3 if and only if it is a contradiction.
Which of the following statements is true?
- a. In set 3, every formula is false in every row of the formulaβs truth table.
- b. In set 2, every formula is true in at least one row and false in at least
- ne row of the formulaβs truth table.
- c. Sets 2 and 3 contain exactly the set of satisfjable formulas.
- d. Two of (a), (b), and (c) are true.
- e. All of (a), (b), and (c) are true.
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Examples
- 1. ((((π β§ π) β (Β¬π )) β§ (π β π)) β (π β (Β¬π )))
- 2. ((((π β§ π) β π ) β§ (π β π)) β (π β π ))
- 3. (π β¨ π) β ((π β§ (Β¬π) β¨ ((Β¬π) β§ π))
- 4. (π β§ (Β¬π))
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How to determine the properties of a formula
- Truth table
- Valuation tree
- Reasoning
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Valuation Tree
Rather than fjll out an entire truth table, we can analyze what happens if we plug in a truth value for one variable.
Β¬T F Β¬F T π β§ T π π β§ F F π β§ π π π β¨ T T π β¨ F π π β¨ π π π β T T π β F Β¬π T β π π F β π T π β π T
We can evaluate a formula by using these rules to construct a valuation tree.
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Example of a valuation tree
Example. Show that (((π β§ π) β (Β¬π )) β§ (π β π)) β (π β (Β¬π ))) is a tautology by using a valuation tree. Suppose π’(π) = T. We put T in for π: (((T β§ π) β (Β¬π )) β§ (T β π)) β (T β (Β¬π )) . Based on the truth tables for the connectives, the formula becomes (((π β (Β¬π )) β§ π) β (Β¬π )). If π’(π) = T, this yields ((Β¬π ) β (Β¬π )) and then T. (Check!). If π’(π) = F, it yields (F β (Β¬π )) and then T. (Check!). Suppose π’(π) = F. We get (((F β§ π) β (Β¬π )) β§ (F β π)) β (F β (Β¬π )) , Simplifjcation yields ((F β (Β¬π )) β§ T) β T and eventually T. Thus every valuation makes the formula true, as required.
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Reasoning about the properties
I found a valuation for which the formula is true. Does the formula have each property below?
- Tautology
YES NO MAYBE
- Contradiction
YES NO MAYBE
- Satisfjable
YES NO MAYBE I found a valuation for which the formula is false. Does the formula have each property below?
- Tautology
YES NO MAYBE
- Contradiction
YES NO MAYBE
- Satisfjable
YES NO MAYBE
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Examples
- 1. ((((π β§ π) β (Β¬π )) β§ (π β π)) β (π β (Β¬π )))
- 2. ((((π β§ π) β π ) β§ (π β π)) β (π β π ))
- 3. (π β¨ π) β ((π β§ (Β¬π)) β¨ ((Β¬π) β§ π))
- 4. (π β§ (Β¬π))
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A logic puzzle
Each of the four cards has a number on one side and a color on the other
- side. How many cards do you have to turn over to test whether this
statement is true: βif a card has an even number on one side, then its
- pposite side is redβ?
Your answer is (a) 0 (b) 1 (c) 2 (d) 3 (e) 4.
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Propositional Logic: Semantics
Alice Gao
Lecture 5, September 21, 2017
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Announcements
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Clickers
- What are they for? Active learning, engagement
- Why do you ask us to answer twice? Peer instruction
- If you choose to be here, please participate!
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Outline for today
- How can we prove that two formulas have the same meaning?
(Logical equivalence)
- Which set of connectives is suffjcient to express all possible formulas?
(Adequate set of connectives)
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Learning goals
Logical equivalence of formulas:
- Prove that the logical equivalence of formulas using truth tables
and/or logical identities.
- Describe strategies to prove logical equivalence using logical identities.
- Translate a condition in a block of code into a propositional logic
formula.
- Simplify code using truth tables and logical identities.
- Determine whether a piece of code is live or dead using truth tables
and logical identities. Adequate set of connectives:
- Prove that a set of connectives is an adequate set for propositional
logic by using truth tables and logical identities.
- Prove that a set of connectives is not an adequate set for
propositional logic.
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Defjnition of logical equivalence
Two formulas π½ and πΎ are logically equivalent if and only if they have the same value under any valuation.
- π½π’ = πΎπ’, for every valuation π’.
- π½ and πΎ must have the same fjnal column in their truth tables.
- (π½ β πΎ) is a tautology.
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Why do we care about logical equivalence?
- Will I lose marks if I provide a solution that is syntactically difgerent
but logically equivalent to the provided solution?
- Do these two circuits behave the same way?
- Do these two pieces of code fragments behave the same way?
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You already know one way of proving logical equivalent. What is it? Theorem: (((Β¬π) β§ π) β¨ π) β‘ (π β¨ π).
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Logical Identities
Commutativity (π½ β§ πΎ) β‘ (πΎ β§ π½) (π½ β¨ πΎ) β‘ (πΎ β¨ π½) Associativity (π½ β§ (πΎ β§ πΏ)) β‘ ((π½ β§ πΎ) β§ πΏ) (π½ β¨ (πΎ β¨ πΏ)) β‘ ((π½ β¨ πΎ) β¨ πΏ) Distributivity (π½ β¨ (πΎ β§ πΏ)) β‘ ((π½ β¨ πΎ) β§ (π½ β¨ πΏ)) (π½ β§ (πΎ β¨ πΏ)) β‘ ((π½ β§ πΎ) β¨ (π½ β§ πΏ)) Idempotence (π½ β¨ π½) β‘ π½ (π½ β§ π½) β‘ π½ Double Negation (Β¬(Β¬π½)) β‘ π½ De Morganβs Laws (Β¬(π½ β§ πΎ)) β‘ ((Β¬π½) β¨ (Β¬πΎ)) (Β¬(π½ β¨ πΎ)) β‘ ((Β¬π½) β§ (Β¬πΎ))
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Logical Identities, contβd
Simplifjcation I (Absorbtion) (π½ β§ T) β‘ π½ (π½ β¨ T) β‘ T (π½ β§ F) β‘ F (π½ β¨ F) β‘ π½ Simplifjcation II (π½ β¨ (π½ β§ πΎ)) β‘ π½ (π½ β§ (π½ β¨ πΎ)) β‘ π½ Implication (π½ β πΎ) β‘ ((Β¬π½) β¨ πΎ) Contrapositive (π½ β πΎ) β‘ ((Β¬πΎ) β (Β¬π½)) Equivalence (π½ β πΎ) β‘ ((π½ β πΎ) β§ (πΎ β π½)) Excluded Middle (π½ β¨ (Β¬π½)) β‘ T Contradiction (π½ β§ (Β¬π½)) β‘ F
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A logical equivalence proof
Theorem: (((Β¬π) β§ π) β¨ π) β‘ (π β¨ π). Proof: (((Β¬π) β§ π) β¨ π) β‘ (((Β¬π) β¨ π) β§ (π β¨ π)) Distributivity ... to be fjlled in ... β‘ (π β¨ π) Simplifjcation I β‘ (π β¨ π) Commutativity QED
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A logical equivalence proof
What is missing from the proof?
- a. (π β¨ π)
- b. (F β§ (π β¨ π))
- c. (π β§ (π β¨ π))
- d. None of these, but I know what it is.
- e. None of these, and thereβs not enough information to tell.
Theorem: (((Β¬π) β§ π) β¨ π) β‘ (π β¨ π). Proof: (((Β¬π) β§ π) β¨ π) β‘ (((Β¬π) β¨ π) β§ (π β¨ π)) Distributivity ... to be fjlled in ... β‘ (π β¨ π) Simplifjcation I β‘ (π β¨ π) Commutativity QED
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A practice problem
βIf it is sunny, I will play golf, provided that I am relaxed.β π‘: it is sunny. π: I will play golf. π : I am relaxed. A few translations:
- 1. (π‘ β (π β π))
- 2. (π β (π‘ β π))
- 3. ((π‘ β§ π ) β π)
Theorem: (π β (π‘ β π)) β‘ ((π‘ β§ π ) β π). Proof: ... to be fjlled in ... QED
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How do you prove non-equivalence?
βIf it snows then I wonβt go to class, but I will do my assignment.β π‘: it snows. π: I will go to class. π: I will do my assignment. 2 possible translations:
- 1. ((π‘ β (Β¬π)) β§ π)
- 2. (π‘ β ((Β¬π) β§ π))
Theorem: ((π‘ β (Β¬π)) β§ π) and (π‘ β ((Β¬π) β§ π)) are not logically equivalent. Which valuation π’ can we use to prove this theorem?
- a. π‘π’ = F, (Β¬π)π’ = F, ππ’ = F
- b. π‘π’ = F, (Β¬π)π’ = T, ππ’ = F
- c. π‘π’ = T, (Β¬π)π’ = T, ππ’ = T
- d. Two of these.
- e. All of these.
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Collected Wisdom
- Try getting rid of β and β.
- Try moving negations inward. Β¬(π β¨ π) β‘ (Β¬π) β§ (Β¬π).
- Work from the more complex side fjrst, BUT
- Switch to difgerent strategies/sides when you get stuck.
- In the end, write the proof in clean βone-side-to-the-otherβ form and
double-check steps.
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A piece of pseudo code
if ( (input > 0) or (not output) ) { if ( not (output and (queuelength < 100) ) ) { π1 } else if ( output and (not (queuelength < 100)) ) { π2 } else { π3 } } else { π4 } When does each piece of code get executed? Let π: input > 0, π£:
- utput,
π: queuelength < 100.
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Simplifying the piece of pseudo code
if ( i or (not u) ) { if ( not (u and q) ) { π1 } else if ( u and (not q) ) { π2 } else { π3 } } else { π4 } Which one of the following is incorrect?
- a. π1 is executed when (π β¨ (Β¬π£)) β§ (Β¬(π£ β§ π)) is true.
- b. π2 is executed when (π β¨ (Β¬π£)) β§ (π£ β§ (Β¬π)) is true.
- c. π3 is executed when ((π β¨ (Β¬π£)) β§ (π£ β§ π)) β§ ((Β¬π£) β¨ π) is true.
- d. π4 is executed when π is false and π£ is true.
- e. All of them are correct.
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A Code Example, contβd
π π£ π (π β¨ (Β¬π£)) (Β¬(π£ β§ π)) (π£ β§ (Β¬π)) T T T T F F π3 T T F T T π1 T F T T T π1 T F F T T π1 F T T F π4 F T F F π4 F F T T T π1 F F F T T π1
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Finding Dead Code
Prove that π2 is dead code. That is, the conditions leading to π2 is logically equivalent to F. (((π β¨ (Β¬π£)) β§ (Β¬(Β¬(π£ β§ π)))) β§ (π£ β§ (Β¬π))) ... to be fjlled in ... β‘ F Simplifying the above condition to F will necessarily use the following logical identities.
- a. Simplifjcation I
- b. Excluded Middle
- c. Contradiction
- d. Two of the above
- e. All of the above
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Finding Live Code
Prove that π3 is live code. That is, the conditions leading to π3 is satisfjable. Theorem: ((π β¨ (Β¬π£)) β§ ((Β¬(Β¬(π£ β§ π))) β§ (Β¬(π£ β§ (Β¬π))))) β‘ ((π β§ π£) β§ π) Proof: ((π β¨ (Β¬π£)) β§ ((Β¬(Β¬(π£ β§ π))) β§ (Β¬(π£ β§ (Β¬π))))) ... to be fjlled in ... β‘ ((π β§ π£) β§ π) QED
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Two pieces of code: Are they equivalent?
Fragment 1: if ( i or (not u) ) { if ( not (u and q) ) { π1 } else if ( u and (not q) ) { π2 } else { π3 } } else { π4 } Fragment 2: if ( (i and u) and q ) { π3 } else if ( (not i) and u ) { π4 } else { π1 }
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Simplifying Code
To prove that the two fragments are equivalent, show that each block of code π1, π2, π3, and π4 is executed under equivalent conditions. Block Fragment 1 Fragment 2 π1 (π β¨ (Β¬π£)) β§ (Β¬(π£ β§ π)) (Β¬(π β§ π£ β§ π)) β§ (Β¬((Β¬π) β§ π£)) π2 (π β¨ (Β¬π£)) β§ (Β¬(Β¬(π£ β§ π))) F β§ (π£ β§ (Β¬π)) π3 (π β¨ (Β¬π£)) β§ (Β¬(Β¬(π£ β§ π))) (π β§ π£ β§ π) β§ (Β¬(π£ β§ (Β¬π))) π4 (Β¬(π β¨ (Β¬π£))) (Β¬(π β§ π£ β§ π)) β§ ((Β¬π) β§ π£)
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The solution to the last logic puzzle
Each of the four cards has a number on one side and a color on the other
- side. How many cards do you have to turn over to test whether this
statement is true: βif a card has an even number on one side, then its
- pposite side is redβ?
Solution: You need to turn over 2 cards. If a card has an even number on
- ne side, then you need to check that its opposite side is red. Also, if a
card is NOT red, you need to check that its opposite side has an ODD number (this is the contrapositive of the given statement). Thus, you need to turn over the second card from the left and the fjrst card from the right.
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Another logic puzzle
A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet three inhabitants: Alice, Rex and Bob. Alice says, βRex is a knave.β Rex says, βitβs false that Bob is a knave.β Bob claims, βI am a knight or Alice is a knight.β Can you determine who is a knight and who is a knave?
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Adequate sets of connectives
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Learning goals
- Prove that a set of connectives is an adequate set for propositional
logic by using truth tables and logical identities.
- Prove that a set of connectives is not an adequate set for
propositional logic.
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Some questions fjrst
- We started propositional logic by learning these connectives Β¬, β§, β¨,
β and β.
- Why did we learn these connectives?
- Using these connectives, can we express every propositional logic
formula that we ever want to write?
- Are there any connectives in this set that are not necessary?
- Are there other connectives that we could defjne and use? Is there
another set of connectives that we should have studied instead?
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Some answers
Is every connective we learned necessary? Nope! Recall that π¦ β π§ β‘ (Β¬π¦) β¨ π§. We donβt need β at all. (We say that β is defjnable in terms of Β¬ and β¨.) Are there other connectives that we could defjne and use? Yep! Letβs take a look.
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Adequate Sets of Connectives
Which set of connectives is suffjcient to express every possible propositional formula? This is called an adequate set of connectives. Any other connective connective is defjnable in terms of the ones in such a set. Theorem 1. {β§, β¨, Β¬} is an adequate set of connectives. Theorem 2. Each of the sets {β§, Β¬}, {β¨, Β¬}, and {β, Β¬} is adequate. Theorem 3. The set {β§, β¨} is not an adequate set of connectives.
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An adequate set to start with
Theorem 1. {β§, β¨, Β¬} is an adequate set of connectives. Hint: use truth tables. How many people know a way to prove this theorem?
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A reduction problem
Theorem 1. {β§, β¨, Β¬} is an adequate set of connectives. Now we can assume that theorem 1 holds. Theorem 2. Each of the sets {β§, Β¬}, {β¨, Β¬}, and {β, Β¬} is adequate. By Theorem 1, the set {β§, β¨, Β¬} is adequate. To prove that {β§, Β¬} is adequate, we need to show that β¨ is defjnable in terms of β§ and Β¬. To prove that {β¨, Β¬} is adequate, we need to show that β§ is defjnable in terms of β¨ and Β¬. To prove that {β, Β¬} is adequate, we need to show that each of β¨ and β§ is defjnable in terms of β and Β¬.
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A non-adequate set
Theorem 3. The set {β§, β¨} is not an adequate set of connectives. Consider any formula which uses only β§ and β¨ as connectives. Assume that every variable in the formula is true. What is the truth value of the formula?
- a. Always true
- b. Always false
- c. Sometimes true and sometimes false
- d. Not enough information to tell
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A non-adequate set
Theorem 3. The set {β§, β¨} is not an adequate set of connectives. Lemma: For any formula which uses only β§ and β¨ as connectives, if every variable in the formula is true, then the formula is true. This lemma means that it is impossible to negate a formula using only β§ and β¨. We can prove the lemma using structural induction.
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