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Published byEustace Barber Modified over 3 years ago

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Proof by Deduction

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Deductions and Formal Proofs A deduction is a sequence of logic statements, each of which is known or assumed to be true A formal proof of a conclusion C is a deduction that ends with C What is known or assumed to be true comes in three variations: –Premises that are assumed to be true –Any statement known to be true (e.g., 1+1 = 2) –Sound rules of inference that introduce logic statements that are true (assuming the statements they are based on are also true)

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Careful Again… A formal proof guarantees that the original statement is a tautology –If any premise is false, the statement is true (P Q is true when P is false) –If all premises are true, we are able to guarantee that the conclusion is true (by the proof derivation) –Thus, the statement is always true (i.e., is a tautology) HOWEVER: –A proof does NOT guarantee that the conclusion is true!

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Valid Proof vs. Valid Conclusion Consider the following proof: If 2 > 3, 2 > 3 3 < 2, then 3 < 2. 1.2 > 3premise 2.2 > 3 3 < 2premise 3.3 < 21&2, modus ponens Is the proof valid? Is the conclusion valid? Yes! Sound argument No! Hence, a valid proof does not guarantee a valid conclusion. What a valid proof does guarantee is that IF the premises are all true, then the conclusion is too.

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Main Rules of Inference A, B |= A B Combination A B |= B Simplification A B |= A Variant of simplification A |= A B Addition B |= A B Variant of addition A, A B |= B Modus ponens B, A B |= A Modus tollens A B, B C |= A C Hypothetical syllogism A B, A |= B Disjunctive syllogism A B, B |= A Variant of disjunctive syllogism A B, A B |= B Cases A B |= A B Equivalence elimination A B |= B A Variant of equivalence elimination A B, B A |= A B Equivalence introduction A, A |= B Inconsistency law

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Reduction of Rules of Inference Using laws, we can also reduce the number of rules of inference –Too many to remember –Too many to program! (Project) Example: modus tollens B A B A B A contrapositive B A modus ponens Easy to show validity of rules with truth tables It turns out that modus ponens plus some simple laws are usually enough, and it can get even simpler…

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1. Exhaustive Truth Proof Show true for all cases –e.g.Prove x 2 + 5 < 20 for integers 0 x 3 0 2 + 5 = 5< 20T 1 2 + 5 = 6< 20T 2 2 + 5 = 9< 20T 3 2 + 5 = 14< 20T –Thus, all cases are exhausted and true –Same as using truth tables when all combinations yield a tautology

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2. Equivalence to Truth 2a.Transform logical expression to T Prove: R S ( R S) R S ( R S) R S R S de Morgan’s law R S R S double negation (R S) (R S) implication (P Q P Q) T law of excl. middle (P P) T Use laws only!

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2. Equivalence to Truth (cont’d) 2b.Transform lhs to rhs or vice versa Prove: P Q P Q Q P Q P Q (P P) Q distributive law (factoring) T Q law of excluded middle Q identity Use laws only!

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3. If and only if Proof Also called necessary and sufficient Since P Q (P Q) (Q P), we can create two standard deductive proofs: –P Q –Q P Transforms proof to a standard deductive proof actually two of them

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3. If and only if Proof: Example Prove: 2x – 4 > 0 iff x > 2. Thus, we can do two proofs: (1) If 2x – 4 > 0 then x > 2. (2) If x > 2 then 2x – 4 > 0. Proof: (1) (2) Similarly, suppose x > 2. Then 2x > 4 and thus 2x – 4 > 0. Note that we could have also converted the lhs to the rhs Proof: 2x – 4 > 0 2x > 4 x > 2 1. 2x – 4 > 0premise 2. 2x > 41, algebraic equiv. 3. x > 22, algebraic equiv.

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Since P Q Q P (by the law of contrapositive), we can prove either side of the equivalence, whichever is EASIER E.g., prove: if s is not a multiple of 3, then s is not a solution of x 2 + 3x – 18 = 0 –Infinitely many non-multiples of 3 and non- solutions to the quadratic equation – HARD! –Let: P = s is a multiple of 3 Q = s is a solution of x 2 + 3x – 18 = 0 –Instead of P Q, we can prove Q P 4. Contrapositive Proof

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Prove: if s is a solution of x 2 + 3x – 18 = 0, then s is a multiple of 3 Proof: The solutions of x 2 + 3x – 18 = 0 are 3 and – 6. Both 3 and – 6 are multiples of 3. Thus, every solution s is a multiple of 3. Note: Using a contrapositive proof helps get rid of the not’s and makes the problem easier to understand indeed, the contrapositive turns this into a simple proof we can do exhaustively. 4. Contrapositive Proof (cont’d)

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Note: P P P F TT F T FT T F Thus, P P F Substituting R S for P, we have R S (R S) F And finally (R S) F ( R S) F implication R S F de Morgan’s R S F double negation 5. Proof by Contradiction

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Thus, we have –R S R S F Hence, to prove R S, we can –Negate the conclusion –Add it as a premise (to R) –Derive a contradiction (i.e. derive F) Any contradiction (e.g., P P) is equivalent to F, so deriving any contradiction is enough 5. Proof by Contradiction (cont’d) Note: in our project we will implement the semantics of our language by programming the computer to do a proof by contradiction

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Prove: the empty set is unique Since P ( P F), we can prove P F instead i.e. We can prove: if the empty set is not unique, then there is a contradiction, or, in other words, assuming the empty set is not unique leads (deductively) to a contradiction. Proof: –Assume the empty set is not unique –Then, there are at least two empty sets 1 and 2 such that 1 2 –Since an empty set is a set and an empty set is a subset of every set, 1 2 and 2 1 and thus 1 = 2 –But now we have 1 2 and 1 = 2 a contradiction! –Hence, the empty set is unique 5. Proof by Contradiction: Example

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