Proof by Deduction

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)

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!

Valid Proof vs. Valid Conclusion Consider the following proof: If 2 > 3, 2 > 3  3 < 2, then 3 < > 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.

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

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…

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

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!

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!

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

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 > x – 4 > 0premise 2. 2x > 41, algebraic equiv. 3. x > 22, algebraic equiv.

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

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)

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

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

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