Table of standard equivalences 30/57 372 TABLES FOR PART I - - PowerPoint PPT Presentation

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/ department of mathematics and computer science

Propositional Logic

Lecture 2 (Chapter 7) September 9, 2016

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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.