CS156: The Calculus of Complete induction (for T PA , T cons ) - PowerPoint PPT Presentation

Induction ◮ Stepwise induction (for T PA , T cons ) CS156: The Calculus of ◮ Complete induction (for T PA , T cons ) Computation Theoretically equivalent in power to stepwise induction, Zohar Manna but sometimes produces more concise proof Winter 2010 ◮ Well-founded induction Generalized complete induction ◮ Structural induction Over logical formulae Chapter 4: Induction Page 1 of 37 Page 2 of 37 Stepwise Induction (Peano Arithmetic T PA ) Example Axiom schema (induction) Prove: F [ n ] : 1 + 2 + · · · + n = n ( n + 1) F [0] ∧ . . . base case 2 ( ∀ n . F [ n ] → F [ n + 1]) . . . inductive step for all n ∈ N . → ∀ x . F [ x ] . . . conclusion ◮ Base case : F [0] : 0 = 0 · 1 2 for Σ PA -formulae F [ x ] with one free variable x . ◮ Inductive step : Assume F [ n ] : 1 + 2 + · · · + n = n ( n +1) , (IH) 2 show To prove ∀ x . F [ x ], the conclusion, i.e., F [ n + 1] : 1 + 2 + · · · + n + ( n + 1) F [ x ] is T PA -valid for all x ∈ N , n ( n + 1) it suffices to show = + ( n + 1) by ( IH ) ◮ base case: prove F [0] is T PA -valid. 2 n ( n + 1) + 2( n + 1) ◮ inductive step: For arbitrary n ∈ N , = 2 assume inductive hypothesis, i.e., ( n + 1)( n + 2) F [ n ] is T PA -valid, = 2 then prove Therefore, F [ n + 1] is T PA -valid. ∀ n ∈ N . 1 + 2 + . . . + n = n ( n + 1) 2 Page 3 of 37 Page 4 of 37

Example: First attempt: ∀ y [ ∀ x . exp 3 ( x , y , 1) = x y ] Theory T + PA obtained from T PA by adding the axioms: � �� � ◮ ∀ x . x 0 = 1 F [ y ] (E0) ◮ ∀ x , y . x y +1 = x y · x We chose induction on y . Why? (E1) ◮ ∀ x , z . exp 3 ( x , 0 , z ) = z (P0) Base case: ◮ ∀ x , y , z . exp 3 ( x , y + 1 , z ) = exp 3 ( x , y , x · z ) (P1) F [0] : ∀ x . exp 3 ( x , 0 , 1) = x 0 For arbitrary x ∈ N , exp 3 ( x , 0 , 1) = 1 (P0) and x 0 = 1 (E0). ( exp 3 ( x , y , z ) stands for x y . z ) Prove that Inductive step: Failure. For arbitrary n ∈ N , we cannot deduce ∀ x , y . exp 3 ( x , y , 1) = x y F [ n + 1] : ∀ x . exp 3 ( x , n + 1 , 1) = x n +1 from the inductive hypothesis is T + PA -valid. F [ n ] : ∀ x . exp 3 ( x , n , 1) = x n Page 5 of 37 Page 6 of 37 Second attempt: Strengthening Inductive step: For arbitrary n ∈ N Assume inductive hypothesis Strengthened property F [ n ] : ∀ x , z . exp 3 ( x , n , z ) = x n · z (IH) prove ∀ x , y , z . exp 3 ( x , y , z ) = x y · z F [ n + 1] : ∀ x ′ , z ′ . exp 3 ( x ′ , n + 1 , z ′ ) = x ′ n +1 · z ′ ↑ note Implies the desired property (choose z = 1) ∀ x , y . exp 3 ( x , y , 1) = x y Consider arbitrary x ′ , z ′ ∈ N : Proof of strengthened property: exp 3 ( x ′ , n + 1 , z ′ ) = exp 3 ( x ′ , n , x ′ · z ′ ) (P1) = x ′ n · ( x ′ · z ′ ) IH F [ n ]; x �→ x ′ , z �→ x ′ · z ′ Again, induction on y = x ′ n +1 · z ′ ∀ y [ ∀ x , z . exp 3 ( x , y , z ) = x y · z (E1) ] � �� � F [ y ] Base case: F [0] : ∀ x , z . exp 3 ( x , 0 , z ) = x 0 · z For arbitrary x , z ∈ N , exp 3 ( x , 0 , z ) = z (P0) and x 0 = 1 (E0). Page 7 of 37 Page 8 of 37

Example: Theory T + Stepwise Induction (Lists T cons ) cons I Axiom schema (induction) T cons with axioms ( ∀ atom u . F [ u ]) ∧ . . . base case Concatenating two lists ( ∀ u , v . F [ v ] → F [cons( u , v )]) . . . inductive step ◮ ∀ atom u . ∀ v . concat ( u , v ) = cons ( u , v ) (C0) → ∀ x . F [ x ] . . . conclusion ◮ ∀ u , v , x . concat (cons( u , v ) , x ) = cons( u , concat ( v , x )) (C1) for Σ cons -formulae F [ x ] with one free variable x . Note: ∀ atom u . F [ u ] stands for ∀ u . (atom( u ) → F [ u ]). To prove ∀ x . F [ x ], i.e., F [ x ] is T cons -valid for all lists x , it suffices to show ◮ base case: prove F [ u ] is T cons -valid for arbitrary atom u . ◮ inductive step: For arbitrary lists u , v , assume inductive hypothesis, i.e., F [ v ] is T cons -valid, then prove F [cons( u , v )] is T cons -valid. Page 9 of 37 Page 10 of 37 Example: Theory T + Example: Theory T + cons II cons III Example: for atoms a , b , c , d , Reversing a list ◮ ∀ atom u . rvs ( u ) = u (R0) concat (cons( a , cons( b , c )) , d ) ◮ ∀ x , y . rvs ( concat ( x , y )) = concat ( rvs ( y ) , rvs ( x )) (R1) = cons( a , concat (cons( b , c ) , d )) ( C 1) = cons( a , cons( b , concat ( c , d ))) ( C 1) Example: for atoms a , b , c , = cons( a , cons( b , cons( c , d ))) ( C 0) rvs (cons( a , cons( b , c )) = rvs ( concat ( a , concat ( b , c ))) ( C 0) concat (cons(cons( a , b ) , c ) , d ) = concat ( rvs ( concat ( b , c )) , rvs ( a )) ( R 1) = cons(cons( a , b ) , concat ( c , d )) ( C 1) = concat ( concat ( rvs ( c ) , rvs ( b )) , rvs ( a )) ( R 1) = cons(cons( a , b ) , cons( c , d )) ( C 0) = concat ( concat ( c , b ) , a ) ( R 0) = concat (cons( c , b ) , a ) ( C 0) = cons( c , concat ( b , a )) ( C 1) = cons( c , cons( b , a )) ( C 0) Page 11 of 37 Page 12 of 37

Example: Theory T + cons IV Prove ∀ x . flat ( x ) → rvs ( rvs ( x )) = x Deciding if a list is flat; � �� � F [ x ] i.e., flat ( x ) is true iff every element of list x is an atom. is T + ◮ ∀ atom u . flat ( u ) cons -valid. (F0) ◮ ∀ u , v . flat (cons( u , v )) ↔ atom( u ) ∧ flat ( v ) (F1) Base case: For arbitrary atom u , Example: for atoms a , b , c , F [ u ] : flat ( u ) → rvs ( rvs ( u )) = u flat (cons( a , cons( b , c ))) = true by F 0 and R0. flat (cons(cons( a , b ) , c )) = false Inductive step: For arbitrary lists u , v , assume the inductive hypothesis F [ v ] : flat ( v ) → rvs ( rvs ( v )) = v (IH) Page 13 of 37 Page 14 of 37 and prove Missing steps: F [cons( u , v )] : flat (cons( u , v )) → rvs ( rvs (cons( u , v ))) rvs ( rvs (cons( u , v ))) = cons( u , v ) ( ∗ ) = rvs ( rvs ( concat ( u , v ))) (C0) = rvs ( concat ( rvs ( v ) , rvs ( u ))) (R1) Case ¬ atom( u ) = concat ( rvs ( rvs ( u )) , rvs ( rvs ( v ))) (R1) = concat ( u , rvs ( rvs ( v ))) (R0) flat (cons( u , v )) ⇔ atom( u ) ∧ flat ( v ) ⇔ ⊥ = concat ( u , v ) (IH), since flat ( v ) by (F1). ( ∗ ) holds since its antecedent is ⊥ . = cons( u , v ) (C0) Case atom( u ) flat (cons( u , v )) ⇔ atom( u ) ∧ flat ( v ) ⇔ flat ( v ) by (F1). Now, show rvs ( rvs (cons( u , v ))) = · · · = cons( u , v ) . Page 15 of 37 Page 16 of 37

Complete Induction (Peano Arithmetic T PA ) Is base case missing? No. Base case is implicit in the structure of complete induction. Axiom schema (complete induction) Note: ( ∀ n . ( ∀ n ′ . n ′ < n → F [ n ′ ] ) → F [ n ]) . . . inductive step ◮ Complete induction is theoretically equivalent in power to � �� � IH stepwise induction. → ∀ x . F [ x ] . . . conclusion ◮ Complete induction sometimes yields more concise proofs. for Σ PA -formulae F [ x ] with one free variable x . Example: Integer division quot (5 , 3) = 1 and rem (5 , 3) = 2 To prove ∀ x . F [ x ], the conclusion i.e., Theory T ∗ PA obtained from T PA by adding the axioms: F [ x ] is T PA -valid for all x ∈ N , ◮ ∀ x , y . x < y → quot ( x , y ) = 0 (Q0) it suffices to show ◮ ∀ x , y . y > 0 → quot ( x + y , y ) = quot ( x , y ) + 1 (Q1) ◮ inductive step: For arbitrary n ∈ N , ◮ ∀ x , y . x < y → rem ( x , y ) = x (R0) assume inductive hypothesis, i.e., F [ n ′ ] is T PA -valid for every n ′ ∈ N such that n ′ < n , ◮ ∀ x , y . y > 0 → rem ( x + y , y ) = rem ( x , y ) (R1) then prove Prove F [ n ] is T PA -valid. (1) ∀ x , y . y > 0 → rem ( x , y ) < y (2) ∀ x , y . y > 0 → x = y · quot ( x , y ) + rem ( x , y ) Best proved by complete induction. Page 17 of 37 Page 18 of 37 Case ¬ ( x < y ): Proof of (1) Then there is natural number n , n < x s.t. x = n + y ∀ x . ∀ y . y > 0 → rem ( x , y ) < y � �� � rem ( x , y ) = rem ( n + y , y ) x = n + y F [ x ] = rem ( n , y ) (R1) Consider an arbitrary natural number x . IH ( x ′ �→ n , y ′ �→ y ) y < Assume the inductive hypothesis since n < x and y > 0 ∀ x ′ . x ′ < x → ∀ y ′ . y ′ > 0 → rem ( x ′ , y ′ ) < y ′ (IH) � �� � F [ x ′ ] Prove F [ x ] : ∀ y . y > 0 → rem ( x , y ) < y . Let y be an arbitrary positive integer Case x < y : rem ( x , y ) = by (R0) x y case < Page 19 of 37 Page 20 of 37