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Proof step 23 from proof ks-RED (previous step)

∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n = 0 ∨ n = 1, ¬ (ap(ks(n)) ∧ ap(ks(n) .1) ∧ ks(n) .1 .1 = K ∧ value(ks(n) .1 .2) ∧ value(ks(n) .2)), ap(ks(n)) ∧ ap(ks(n) .1) ∧ ap(ks(n) .1 .1) ∧ ks(n) .1 .1 .1 = S, v = ks(n) .2, v0 = ks(n) .1 .2, v4 = ks(n) .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v) ⊦ 〈tn0 := some(v4 _ v _ (v0 _ v))〉〈tn := tn0 ; if tn = none ∧ ap(ks(n)) then { RED1(ks(n) .1; tn) ; if tn ≠ none then tn := some(tn .t _ ks(n) .2) else if value(ks(n) .1) then { RED1(ks(n) .2; tn) ; if tn ≠ none then tn := some(ks(n) .1 _ tn .t) } }〉〈if tn ≠ none then RED(tn .t; t0) else t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 22:
∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n = 0 ∨ n = 1, ¬ (ap(ks(n)) ∧ ap(ks(n) .1) ∧ ks(n) .1 .1 = K ∧ value(ks(n) .1 .2) ∧ value(ks(n) .2)), ap(ks(n)) ∧ ap(ks(n) .1) ∧ ap(ks(n) .1 .1) ∧ ks(n) .1 .1 .1 = S, v = ks(n) .2, v0 = ks(n) .1 .2, v4 = ks(n) .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v) ⊦ 〈tn := some(v4 _ v _ (v0 _ v))〉〈if tn = none ∧ ap(ks(n)) then { RED1(ks(n) .1; tn) ; if tn ≠ none then tn := some(tn .t _ ks(n) .2) else if value(ks(n) .1) then { RED1(ks(n) .2; tn) ; if tn ≠ none then tn := some(ks(n) .1 _ tn .t) } }〉〈if tn ≠ none then RED(tn .t; t0) else t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]