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Proof step 25 from proof ks-RED (previous step)

∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), t1 = S _ t2, ap(ks(1)), ap(ks(1) .1), ¬ (value(t2) ∧ value(t3) ∧ value(ks(1) .2)), ks(1) .1 = t1 _ t3 ⊦ 〈RED1(ks(1) .1; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(ks(1) .1) then { RED1(ks(1) .2; 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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), ks(1) .1 = S _ t2 _ t3, ap(ks(1)), ¬ (value(t2) ∧ value(t3) ∧ value(ks(1) .2)) ⊦ 〈RED1(S _ t2 _ t3; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(S _ t2 _ t3) then { RED1(ks(1) .2; 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 : ([CSIMPSEQ : ⊦ ap(t _ t0)])]]