proof by deduction. deductions and formal proofs a deduction is a sequence of logic statements, each...
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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 ConclusionConsider the following proof:
If 2 > 3, 2 > 3 3 < 2, then 3 < 2 .
1. 2 > 3 premise
2. 2 > 3 3 < 2 premise
3. 3 < 2 1&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 InferenceA, 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, AB |= B Modus ponens
B, AB |= A Modus tollens
AB, BC |= AC Hypothetical syllogism
A B, A |= B Disjunctive syllogism
A B, B |= A Variant of disjunctive syllogism
AB, AB |= B Cases
AB |= AB Equivalence elimination
AB |= BA Variant of equivalence elimination
AB, BA |= AB 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 tollensBA BA
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 x2 + 5 < 20 for integers 0 x 3
02 + 5 = 5 < 20 T12 + 5 = 6 < 20 T22 + 5 = 9 < 20 T32 + 5 = 14 < 20 T
– 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 (PQ PQ)
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 versaProve: 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 rhsProof: 2x – 4 > 0 2x > 4 x > 2
1. 2x – 4 > 0 premise2. 2x > 4 1, algebraic equiv.3. x > 2 2, 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 x2 + 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 x2 + 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 x2 + 3x – 18 = 0, then s is a multiple of 3
Proof: The solutions of x2 + 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
T T F T
F T 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 uniqueSince 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