Functional Programming Lecture 13 - induction on lists

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Functional ProgrammingLecture 13 - induction
on lists
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Proof by induction on Nat

• Nat is an inductively defined set
• base case 0 Nat,
• ind. case if n Nat, then n1 Nat.
• extremal nothing else belongs to Nat.
• To prove a property of Nat, P(x), we use the
  Principle of Induction
• base case prove P(0),
• ind. case Prove P(n1), assuming P(n).

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Proof by induction on Lists

• Lists are inductively defined sets
• base case a,
• ind. case if xs a, and xa, then
  (xxs) a.
• extremal nothing else belongs to a.
• To prove a property of a, P(xs), we use the
  Principle of (Structural) Induction
• base case prove P(),
• ind. case Prove P(xxs), assuming P(xs).
• The format of the proof is extremely important.
  Best illustrated by examples.

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• sumlist 0 s.0
• sumlist (xxs) x sumlist xs s.1
• double d.0
• double (xxs) (2x) double xs d.1
• Theorem For all finite xs.
• sumlist (double xs) 2 (sumlist xs)
• Proof By induction on xs.
• Base Case Prove sumlist (double ) 2
  (sumlist ).
• l.h.s. sumlist (double ) sumlist by
  d.0
• 0
  by s.0
• r.h.s. 2 (sumlist ) 2 0
  by s.0
• 0
  by arith.
• Ind Case Assume sumlist (double xs ) 2
  (sumlist xs). Prove sumlist (double xxs ) 2
  (sumlist xxs).
• l.h.s. sumlist (double xxs )
• sumlist ((2x) double xs)
  by d.1
• (2x) sumlist (double xs)
  by s.1

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Important Points

• Justify every step.
• Clearly state assumption.
• If you do not use the assumption in a step, then
  you are not doing a proof by induction!
• Work on the l.h.s., or the r.h.s., or both.

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• xs xs
  .0
• xxs ys x(xs ys) .1
• Theorem For all finite xs. xs xs.
• ( is a right identity for )
• Proof By induction on xs.
• Base Case Prove .
• l.h.s.
  by .0
• Ind Case Assume xs xs.
• Prove xxs xxs.
• l.h.s. xxs
• x(xs )
  by .1
• xxs
  by ass.
• QED

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• xs xs
  .0
• xxs ys x(xs ys) .1
• Theorem For all finite xs,yz,zs.
• xs (ys zs) (xs ys) zs
• ( is associative)
• Proof By induction on xs.
• Base Case Prove (ys zs) ( ys)
  zs
• l.h.s. (ys zs) ys zs
  by .0
• ( ys)
  zs by .0
• Ind Case Assume xs (ys zs) (xs ys)
  zs
• Prove xxs (ys zs) (xxs ys) zs
• l.h.s. xxs (ys zs) x(xs (yszs))
  by .1
  
• x((xs ys)zs)
  by ass.
• x(xsys) zs
  by .1
• (xxs ys) zs
  by .1

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• member String -gt Char -gt Bool
• member y False
  m.0
• member (xxs) y (xy) member xs y
  m.1
• False x x
  or.0
• True x True
  or.1

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• Theorem For all finite xs, ys, and elements z.
• member (xs ys ) z member xs z member ys
  z.
• Proof By induction on xs.
• Base Case Prove
• member ( ys ) z member z member ys
  z.
• l.h.s. member ( ys ) z member ys z by
  .0
• r.h.s. member z member ys z
• False member ys z
  by m.1
• member ys z
  by or.0
• Ind Case Assume member (xs ys ) z member xs
  z member ys z. Prove
• member (xxs ys ) z member (xxs) z
  member ys z.
• l.h.s. member (xxs ys ) z
• member (x(xs ys))z
  by .1

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An Exercise

• prove that is associative.
• Question Why dont we need induction?

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• rev
  rev.0
• rev xxs rev xs x rev.
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• Theorem For all finite xs. rev (rev xs) xs
• Proof By induction on xs.
• Base Case Prove rev (rev )
• l.h.s. rev (rev ) rev
  by rev.0
• 
  by rev.0
• Ind Case Assume rev (rev xs) xs. Prove
• rev (rev xxs) xxs.
• l.h.s. rev (rev xxs) rev (rev xs x)
  by rev.1
• rev (x) rev
  (rev xs) by ??

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• rev
  rev.0
• rev xxs rev xs x rev.
  1
• Theorem For all finite xs,ys.
• rev (xs ys) (rev ys) (rev xs)
• Proof By induction on xs.
• Base Case Prove rev ( ys) (rev )
  (rev ys).
• l.h.s. rev ( ys) rev ys
  by .0
• rev ys
  by .0
• (rev ) (rev
  ys) by rev.0
• Ind Case Assume rev (xs ys) (rev ys) (rev
  xs).
• Prove rev (xxs ys) (rev ys) (rev xxs).
• l.h.s. rev (xxs ys) rev (x(xsys))
  by .1
• rev(xsys) x
  by rev.1