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Proof step 7 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), tn0 = some(ks(n) .1 .2) ⊦ 〈RED(tn0 .t; t0)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), ks(1) .1 .1 = K, value(ks(1) .2), value(ks(1) .1 .2), ap(ks(1) .1), ap(ks(1)) ⊦ 〈RED(ks(1) .1 .2; t0)〉 t0 = K _ K

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ odd(1)], [CSIMPSEQ : ⊦ ¬ even(1)], [CSIMPSEQ : ⊦ some(t) .t = t], [CSIMPSEQ : ⊦ ¬ ap(K)], [CSIMPSEQ : ⊦ ks(0) = K])]]