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Proof step 61 from proof ks-RED (previous step)

∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), tn1 = none ⊦ 〈skip〉〈n := 1 ; tn0 := tn1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else { RED1(K; tn0) ; if tn0 ≠ none then tn0 := some(ks(n) .1 _ tn0 .t) }〉〈if tn0 ≠ none then RED(tn0 .t; t0) else t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Heuristic symbolic execution applied rule skip right

And obtained the following new goals

Proof step 60:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), tn1 = none ⊦ 〈n := 1〉〈tn0 := tn1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else { RED1(K; tn0) ; if tn0 ≠ none then tn0 := some(ks(n) .1 _ tn0 .t) }〉〈if tn0 ≠ none then RED(tn0 .t; t0) else t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]