CMPSCI 250 Discussion #3: Practicing Proofs

David Mix Barrington

26 September 2007

In this discussion we applied our proof methods, for both the propositional and predicate calculi, to prove some statements about numbers and strings.

First, recall the definitions of a string over a fixed alphabet, the concatenation of two strings, and the reversal w^R of a string w (which is w written backwards). We will assume the following two facts, which we aren't prepared to prove yet:

• Fact I: ∀w: (w^R)^R = w
• Fact II: ∀u:∀v: (uv)^R = v^Ru^R

A palindrome is a string w such that w^R = 2. Examples the empty string λ, "hannah", and a statement attributed to Napoleon, "Able was I ere I saw Elba", which is a palindrome as long as we ignore capitalization. We'll write Pal(w) to mean "w is a palindrome".

One string is a prefix of another if you can make the second by concatenating some string onto the first -- using quantifiers:

Pre(u,v) ↔ ∃w:(uw=v)

Writing Exercises:

1. We begin by using the quantifier rules to prove some statements about natural numbers as in Discussion #1. We define Even(x) to mean ∃y: x = 2y and Odd(x) to mean ∃y: x = 2y+1. Prove:

   ∀x:forall;y: [Odd(x) ∧ Odd(y)] → [Even(x+y) ∧ Odd(xy)]

   Since we want to prove a universal statement, we start by letting a and b we arbitrary naturals. We assume Odd(a) and Odd(b) to start a direct proof of the implication [Odd(a) ∧ Odd(b)] → [Even(a+b) ∧ Odd(ab)].

   By the definition of Odd, we have ∃c:a = 2c+1 and ∃d: b = 2d+1. By Instantiation, let c and d be naturals such that a = 2c+1 and b = 2d+1. By arithmetic, a+b = 2c + 2d + 2 = 2(c+d+1). By Existence, letting e = c+d+1, we get that ∃e: a+b = 2e which is the definition of Even(a+b). By more arithmetic, ab = (2c+1)(2d+1) = 4cd + 2c + 2d + 1 = 2(2cd+c+d) + 1. Using Existence with f = 2cd+c+d, we have ∃f: ab = 2f+1 which is the definition of Odd(ab).

   Assuming Odd(a) ∧ Odd(b), we have proved Even(a+b) ∧ Odd(ab), so we have proved the desired implication by direct proof.

   Since a and b were arbitrary, by Generalization we may conclude ∀x:∀y:[Odd(x) ∧ Odd(y)] → [Even(x+y) ∧ Odd(ab)].

2. Now using the Pal and Pre predicates defined above, prove

   ∀x:∃y: Pre(x,y) ∧ Pal(y)

   1. Let a be an arbitrary string.
   2. Let b be aa^R. (Note: This is the creative step -- we have to convince ourselves that this is the string we want.)
   3. Letting c be a^R, note that ac = b and hence ∃c: ac = b by Existence.
   4. By the definition of Pre, we have that Pre(a,b).
   5. Since (aa^R)^R = (a^R)^Ra^R by Fact II which equals aa^R by Fact I, we know that Pal(aa^R) and hence Pal(b).
   6. Using Existence on b, we have that ∃y:Pre(a,y) ∧ Pal(y).
   7. Since a was arbitrary, by Generalization we have ∀x:∃y: Pre(x,y) ∧ Pal(y).
3. Prove ∃u:∀v:∃w: (v = (uw)^R), where again all variables denote strings.

   This one is tricky in that the first step of the formal proof requires us to decide what u is going to be. But note that u is going to have the property that for any v, we can find a w to make v = (uw)^R true. If v happens to be empty, uw will have to be empty and thus u will have to be empty.

   1. We need to prove ∀v:∃w: v = (λw)^R.
   2. Let v be arbitrary.
   3. Let w be v^R. Note that λw = w.
   4. By Fact I, w^R = (v^R)^R = v.
   5. By Existence, we have that ∃w: v = (λw)^R.
   6. Since v was arbitrary, we have that ∀v:∃w: v = (λw)^R.
   7. By Existence, letting u be λ, we have ∃u:∀v:∃w: v = (uw)^R, as desired.
4. Prove that Pre is a partial order on the set of all strings. To prove antisymmetry, use the fact that every string w has a length |w|, the fact that |uv| = |u| + |v|, and arithmetic on naturals.
   • Pre is reflexive: We must prove ∀x: Pre(x,x). Because x = xλ, by Existence we have ∃w: xw = x, which is the definition of Pre(x,x).
   • Pre is antisymmetric: We must prove ∀u:∀v:[Pre(u,v) ∧ Pre(v,u)] → (u=v). Let u and v be arbitrary strings and assume Pre(u,v) and Pre(v,u). By Instantiation on the definitions of Pre(u,v) and Pre(v,u), let w and x be strings such that uw = v and vx = w. By the given rules about lengths, |v| = |uw| = |u| + |w| and |u| = |vx| = |v| + |x|. Since |w| and |x| are naturals and cannot be negative, we must have |w| = |x| = 0 so that w = x = λ. But then u = uλ = v. We have proved that [Pre(u,v) ∧ Pre(v,u)] → (u=v), and since u and v were arbitrary we have proved antisymmetry.
   • Pre is transitive: We must prove ∀u:∀v:∀w: [Pre(u,v) ∧ Pre(v,w)] → Pre(u,w). So let u, v, and w be arbitrary strings and assume Pre(u,v) and Pre(v,w). By Instantiation on the definition of Pre, let x and y be strings such that ux = v and vy = w. Then by substitution, w = vy = (ux)y. Letting z = xy, we have w = uz and hence ∃z: uz = w by Existence. This is the definition of Pre(u,w). We have proved the desired implication by direct proof, and since u, v, and w were arbitrary we can conclude the quantified statement by Generalization.

Last modified 26 September 2007