More Formal Foundations: Proof Power

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More Formal FoundationsProof Power

• Helpful Reading
• Rosen 3.1 - 3.2

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When we last left our heroes

• Groundwork for the propositional calculus
• Demonstrating logical equivalence
• Turning predicates into propositions
• Specifying a value for an unknown
• Quantifying a predicate over the universe of
  discourse

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Today Payday

• Today, you learn what all this stuff weve been
  discussing means
• Constructing an argument step by step to prove
  theorems (a theorem is any statement that can be
  proved true)
• Generally these statements are in the space of
  predicates and not propositions
• proofs over the whole universe of discourse
• proofs for quantified predicates

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The Nature of Proof

• We usually want to prove implications (for
  example, If n is odd, then n2 is also odd.) --
  usually truths that depend on other truths
• We try to show that knowing the if part is true
  leads logically to certain truths that make up
  the then part
• If each step is logically sound, then you must
  accept the end result

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The Direct Proof

• Proving that the statement pq is true by showing
  that p logically leads to q
• Predicate calculus applies here for true
  correctness
• Example If n is odd, then n2 is odd.
• P(n) n is odd
• Q(n) n2 is odd
• The statement is really "n (P(n) Q(n)) where
  the universe of discourse is (nonnegative)
  integers

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Direct Proof Example

• Let n be an element in the universe of discourse
• Assume n is odd, show that its square is odd
• n odd there is an integer m so that n 2m 1
• So, n2 (2m 1)(2m 1) 4m2 4m 1 2(2m2
  2m) 1
• Since m is an integer, 2m2 2m is also an
  integer
• Thus, there is an integer p such that n2 2p
  1, and so n2 is odd

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Indirect Proof

• Since pq is logically equivalent to ØqØp, you
  can prove the former by directly proving the
  latter this is called an indirect proof
• Lets prove the statement we just examined
  indirectly -- in other words, show that "n (ØQ(n)
  ØP(n))

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Indirect Proof Example

• Let n be an element in the universe of discourse
• Assume n2 is not odd
• Thus there must exist an integer m such that n2
  2m
• Thus n Ö (2m) Ö (2) Ö (m)
• But Ö (2) is not an integer, so it follows that m
  must have at least one factor of 2 in it, and
  thus there is some integer p so that m2p, and
  further, Ö(p) must also be an integer
• n 2Ö(p) , and since Ö(p) is an integer, n is
  even

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Proof by Contradiction

• Show that pq is true by initially assuming that
  pÙØq is true
• Contradiction arises because Øq logically leads
  to Øp since pq is actually true, and thus we get
  pÙØp
• Thus pÙØq is false, and its negation must be
  true, but its negation is exactly equivalent to
  the original implication
• In general contradiction proof, assume the
  negation of what you are trying to prove

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Proof By Contradiction Exmpl.

• Prove that Ö (2) is not a rational number
• Assume that it is, and so it can be expressed as
  a lowest-terms fraction of two integers a/b (a
  and b have no common factors)
• So 2 a2/b2, and 2b2 a2, so a2 is even
• Since a2 is even, a is even (note that this is
  just the contrapositive of what we proved
  earlier)
• Thus, a 2c where c is some integer, and
  therefore 2b2 4c2 2(2c2)
• So b is also even, thus a and b are both
  divisible by 2, but the fraction was in lowest
  terms, so there is a contradiction -- assumption
  was false

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Other Proofs on Implications

• pq is always true if q is always true, or if p
  is always false
• Suggests two other forms of proof
• Trivial proof Ignore p and show q is always true
• Vacuous proof Ignore q and show that p is always
  false
• These are useful as steps in a sub-proof or
  nested proof

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The Inductive Proofs

• Used to show truth over a universe of discourse
  consisting of the integers larger than some base
  number
• Example Prove that for all integers n, the sum
  of the integers from 1 to n is n(n1)/2
• Let P(m) be Sum of integers from 1 to m is
  m(m1)/2
• Want to show "n1 P(n)

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Weak Induction

• Most common form of inductive proof
• To show P(n) true for all nc
• Show P(c), called the basis step
• Show for any arbitrary m, P(m)P(m1), called the
  inductive step
• Since it works for c, it must work for (c1), and
  since it works for (c1), it must work for (c2),
  and so on and so forth...

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Weak Induction Example

• Prove that for any positive integer n, a 2n x 2n
  chessboard with one square removed can be tiled
  entirely with three-square, L-shaped tiles
• Basis Prove P(1). A 2 x 2 chessboard with one
  square removed looks exactly like an L-shaped
  tile
• Inductive Step Assume P(n), show P(n1). For
  the 2n1 x 2n1 chessboard, split it into four
  quadrants each of size 2n x 2n
• For the quadrant with the missing square,
  inductive hypothesis P(n) applies -- we assume we
  can tile it
• For the quadrants without the missing square,
  note that we can temporarily remove the corners
  to make another L, tile whats left, and then
  tile the missing corners

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Strong Induction

• To show P(n) true for all nc
• Show P(c) and any additional cases for (c1),
  (c2), (c3), etc. as needed
• Show P(c) P(c1) P(n-1)P(n)
• Most often used when P(m) does not lead
  intuitively to a proof for P(m1), or when such a
  proof has holes
• Main idea is the same the fact that it worked
  before works now

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Strong Induction Example

• Prove that postage of 12 cents or higher can be
  made using only 4-cent and 5-cent stamps
• Multiple basis steps
• P(12) - Three 4-cent stamps
• P(13) - Two 4-cent stamps and a 5-cent stamp
• P(14) - A 4-cent stamp and two 5-cent stamps
• P(15) - Three 5-cent stamps
• Inductive step for n 16
• Simply add a four-cent stamp to the way you
  solved the problem in P(n-4)
• Can also be done with weak induction, but the
  proof is much more difficult

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Existence Proofs

• Finally, lets talk about how to prove statements
  of the form x P(x)
• Remember that all you have to do is show that
  there is an item that fulfills the predicate
• Two proof styles
• Constructive prove x P(x) by identifying such
  an x explicitly
• Nonconstructive prove that there must be an x
  fulfilling x P(x) without actually saying what
  that x is

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Nonconstructive Proof Ex.

• Show that for any integer n, there must be a
  prime number larger than n
• If P(n) n is prime and G(m, n) m is larger
  than n, the logical expression of this statement
  is "n x (P(x) Ù G(x, n))
• Proof Consider the integer n! 1
• Since every integer has at least one prime
  factor, at least one prime divides this integer
• But division by any number between 2 and n
  results in a remainder of 1
• Thus, any prime factor must be larger than n