Mathematical Induction

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1 Mathematical Induction Definition of Structures A powerful proof technique for proving theorems about recursively defined data structures is mathematical induction. It goes beyond first-order logic. It can be described in higher-order logic, but because of its importance special proof techniques have been developed. Example for data structures natural numbers (constructors: 0 and s) lists (constructors: nil and cons) stacks (constructors: emptystack and makestack) Mechanised Deduction, c M. Kerber,

2 Natural Numbers Peano axioms: 1. N(0) 2. n N(n) N(s(n)) 3. n 0 = s(n) 4. n, m s(n) = s(m) n = m 5. P (P (0) ( n P (n) P (s(n)))) n P (n) Last axiom is higher-order. Abbreviate axioms by data structure with constructors: 0 and s Mechanised Deduction, c M. Kerber,

3 Natural Numbers (Cont d) Definition of functions. Example +: 1. x x + 0 = x 2. x, y x + s(y) = s(x + y) How to prove properties of + like x 0 + x = x? Show: = 0 (holds by first case of definition of + with [0/x]) 2. x 0 + x = x 0 + s(x) = s(x) (Assume 0 + x = x, calculate 0 + s(x) = s(0 + x) by second case of definition of + with [0/x], [x/y]. By assumption 0 + x = x, hence s(0 + x) = s(x), that is, 0 + s(x) = s(x)) Mechanised Deduction, c M. Kerber,

4 Example Lists constructors: nil and cons. Definition of functions. Example append: 1. x append(nil, x) = x 2. e x, y append(cons(e, x), y) = cons(e, append(x, y)) How to prove properties of append like x append(x, nil) = x? Show: 1. append(nil, nil) = nil (holds by first case of definition of append with [nil/x]) 2. e x append(x, nil) = x append(cons(e, x), nil) = cons(e, x) (Assume append(x, nil) = x, calculate (for arbitrary e), append(cons(e, x), nil) by second case of definition of append simplifies to cons(e, append(x, nil)). Since append(x, nil) = x by assumption, this further simplifies to cons(e, x) q.e.d.) Mechanised Deduction, c M. Kerber,

5 Induction Proofs In general show for a property (like x P (x) 0 + x = x) Base case(s): Show that the property holds for any of the base cases. (i.e., for N show P (0)) Step case(s): Assume that the property holds (i.e., for N assume P (x)) up to a certain level and show that by the application of one constructor the property is conserved (i.e., for N prove P (s(x)) under the assumption P (x)). Some problems with mathematical induction: Find an appropriate induction variable Find an appropriate induction schema Generalisation Find appropriate lemmata Mechanised Deduction, c M. Kerber,

6 Find Appropriate Induction Variable Example: Prove x, y, z x + (y + z) = (x + y) + z by induction over x, over y, or over z? Induction over x Base case: y, z (0 + y) + z = 0 + (y + z) need lemma: x 0 + x = x Induction over z Base case: x, y (x+y)+0 = x+(y+0) trivial Step case: x, y, z (x+y)+z = x+(y +z) (s(x)+y)+z = s(x)+(y +z) need lemma: x, y s(x) + y = s(x + y) Step case: x, y, z (x+y)+z = x+(y +z) (x+y)+s(z) = x+(y +s(z)) proved by just three applications of definition of + Mechanised Deduction, c M. Kerber,

7 Find an Appropriate Induction Schema Example: Define odd and even as: even(0), x even(x) odd(s(x)), and x odd(x) even(s(x)). (additional problem: Is this a definition at all?) Show: x, y even(x) even(y) even(x + y). How?: Base case: x even(x) even(0) even(x + 0) Step case: x, y (even(x) even(y) even(x + y)) (even(x) even(s(s(y))) even(x + s(s(y)))) Why correct? Mechanised Deduction, c M. Kerber,

8 Generalisation Example: Try to prove x x + (x + x) = (x + x) + x Induction over x: Base case: 0 + (0 + 0) = (0 + 0) + 0 simple Step case: x x + (x + x) = (x + x) + x s(x) + (s(x) + s(x)) = (s(x) + s(x)) + s(x) Under the assumption x + (x + x) = (x + x) + x it is not possible to prove s(x) + (s(x) + s(x)) = (s(x) + s(x)) + s(x) directly, but as seen on Slide 110 it is possible to prove the generalisation x, y, z (x + y) + z = x + (y + z) (x + y) + s(z) = x + (y + s(z)) directly. Mechanised Deduction, c M. Kerber,

9 Finding Lemmata Example: Define half as: half (0) = 0 half (1) = 0 x half (s(s(x))) = s(half (x)) Show half (x + x) = x (base case trivial) Step case: x half (x + x) = x half (s(x) + s(x)) = s(x) proved by: half (s(x) + s(x)) Def = + half (s(s(x) + x)) Lemma = half (s(s(x + x))) Def half = s(half (x + x)) IndHyp = s(x) How to speculate the lemma if it is not given? Mechanised Deduction, c M. Kerber,

10 The Waterfall Model of Inductive Thm Proving Top level strategy of NQTHM as waterfall. Top = initial clauses. Remove clauses C repeatedly from the top and pour it over the waterfall (until Top empty) by trying procedures in order, which can be described as tactics. If no procedure succeeds C comes into the pool. When the top is empty, a clause C is selected from the pool for induction. An induction schema is applied to this clause and the result is pumped to the top. When top and pool empty, initial conjecture proved. Mechanised Deduction, c M. Kerber,

11 Rippling Although there are particular problems about the proving theorem by mathematical induction, there is a special structural information in the step case between the induction hypothesis P (n) and the induction conclusion P ( s( n ) ). Essentially the constructor s can be considered as an interference factor which has to be moved to a position where it doesn t harm. Example: Use induction hypothesis x + (y + z) = (x + y) + z to prove induction conclusion x + (y + s( z ) ) = (x + y) + s( z ) Rippling out to move difference to the top of the term: over x + ( s( y + z ) ) = s( (x + y) + z ) to s( x + (y + z) ) = s( (x + y) + z ) Here induction hypothesis applicable Mechanised Deduction, c M. Kerber,

12 Rippling (Cont d) General idea of rippling: Restrict search space by applying rules in a restricted way only, e.g. x + s( y ) s( x + y ) even( s(s( x )) ) even(x) Conclusion Inductive theorem proving requires more basic decisions than first-order theorem prover (e.g., with a bad induction schema no proof can be derived) In inductive theorem proving structure can be exploited by derivation of induction conclusion from induction hypothesis Mechanised Deduction, c M. Kerber,

Full and Complete Binary Trees

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