CSE 311 Lecture 09: Proof Strategies

Topics

Predicate logic proofs

A review and continuation of Lecture 08.

Natural language proofs

From formal proofs to natural language proofs.

Proof strategies

Proof by contrapositive, counterexamples, and proof by contradiction.

Inference rules for quantifiers

Predicate logic proofs can use …

Predicate logic inference rules

Applied to whole formulas only.

Predicate logic equivalences

Even on subformulas.

Propositional logic inference rules

Applied to whole formulas only.

Propositional logic equivalences

Even on subformulas.

Predicate logic proofs can also use domain properties

Prove that there is an even prime number: $\exists x. \text{Even}(x) \wedge \text{Prime}(x)$.

We give an explicit logic definition of Even but use a black-box definition of Prime because the proof won’t need to break it down further.

An equal example to demonstrate Elim $\forall$ and Intro $\exists$

Prove that $\forall y. \exists z. y = z$ follows from $\forall x. x = x$.

A square example to demonstrate dependencies

Prove that the square of every even number is even: $\forall x. \text{Even}(x) \rightarrow \text{Even}(x^2)$.

Why list dependencies? To avoid incorrect proofs.

Over the integer domain: $\forall x. \exists y. y\geq x$ is True but $\exists y. \forall x. y\geq x$ is False.

Natural language versus (predicate) logic proofs

We often write proofs in English rather than as fully formal proofs.

They are easier for people to read.

(But theorem provers prefer fully formal proofs. :)

English proofs follow the structure of the corresponding formal proofs.

Formal proof methods help to understand how proofs work in English.

And they give clues for how to produce the proofs in English.

The not so odd example in English

Prove that there is an even integer.

The square example in English

Prove that the square of every even number is even.

An odd square example in English

Prove that the square of every odd number is odd.

Proof

Let $b$ be an arbitrary odd number.

Then, $b = 2c + 1$ for some integer $c$ (depending on $b$).

Therefore, $b^2 = (2c + 1)^2 = 4c^2 + 4c + 1 = 2(2c^2+2c) + 1$.

Since $2c^2+2c$ is an integer, $b^2$ is odd.

The statement follows since $b$ was arbitrary. $\qed$

A rational example in English

A real number $x$ is rational iff there exist integers $p$ and $q$ with $q\neq0$ such that $x=p/q$.

Prove: “If $x$ and $y$ are arbitrary rational numbers then $xy$ is rational.”

Proof

By the definition of rational, $x = a/b$ for some integers $a$, $b$, where $b\neq0$, and $y = c/d$ for some integers $c$, $d$, where $d\neq0$.

Multiplying, we get that $xy = (ac)/(bd)$.

Since $b$ and $d$ are both non-zero, so is $bd$; furthermore, $ac$ and $bd$ are integers.

It follows that $xy$ is rational, by definition of rational.

Benefits of English proofs

%a = add %i, 1
%b = mod %a, %n
%c = add %arr, %b
%d = load %c
%e = add %arr, %i
store %e, %d

arr[i] = arr[(i+1) % n];

Proof by contrapositive

If we assume $\neg q$ and derive $\neg p$, then we have proven that $\neg q \rightarrow \neg p$, which is equivalent to proving $p \rightarrow q$.

Counterexamples

To disprove $\forall x. P(x)$, prove $\exists x. \neg P(x)$.

Works by DeMorgan’s Law: $\neg \forall x. P(x)\equiv \exists x.\neg P(x)$.

All we need to do is find an $x$ for which $P(x)$ is false.

This $x$ is called a counterexample.

Example: disprove that “Every prime number is odd”.

2 is a prime number that is not odd.

Proof by contradiction

If we assume $p$ and derive $\F$ (a contradiction), then we have proven $\neg p$.

An example proof by contradiction

Prove that “No integer is both even and odd.”

English proof: $\neg\exists x. \text{Even}(x)\wedge\text{Odd}(x) \equiv \forall x. \neg (\text{Even}(x)\wedge\text{Odd}(x))$.

Proof by contradiction

Let $x$ be an arbitrary integer and suppose that it is both even and odd.

Then $x=2a$ for some integer $a$ and $x=2b+1$ for some integer $b$.

Therefore $2a=2b+1$ and hence $a=b+\frac{1}{2}$.

But two integers cannot differ by $\frac{1}{2}$ so this is a contradiction.

Therefore no integer is both even and odd. $\qed$

Fun strategy: proof by computer

Use an automated theorem prover:

; No integer is both even and odd.

(define-fun even ((x Int)) Bool
  (exists ((y Int)) (= x (* 2 y))))

(define-fun odd ((x Int)) Bool
  (exists ((y Int)) (= x (+ (* 2 y) 1))))

(define-fun claim () Bool
  (not (exists ((x Int)) (and (even x) (odd x)))))

(assert (not claim)) ; proof by contradiction

(check-sat)

While this example works, proofs of arbitrary formulas in predicate logic cannot be automated. But interactive theorem provers can still help by checking your formal proof and filling in some low-level details for you.

Fun fact: counterexamples & contradiction in verification

Automated verifiers work by counterexample and contradiction proofs.

Recall that program verification involves proving that a program $P$ satisfies a specification $S$ on all inputs $x$: $\forall x. p(x) \rightarrow s(x)$, where $p$ and $s$ are formulas encoding the semantics of $P$ and $S$.

The program verifier sends the formula $\exists x. p(x) \wedge \neg s(x)$ to the prover.

$\neg \forall x. p(x) \rightarrow s(x) \equiv \exists x. \neg (p(x) \rightarrow s(x)) \equiv \exists x. \neg (\neg p(x) \vee s(x))\equiv \exists x. p(x) \wedge \neg s(x)$.

If the prover finds a counterexample, we know the program is incorrect.

The counterexample is a concrete input (test case) on which the program violates the spec.

If no counterexample exists, we know the program is correct.

Because this is proof by contradiction! The prover assumed $\exists x. p(x) \wedge \neg s(x)$ and arrived at false (“unsat”).