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Proof step 6 from proof ks-RED (previous step)

∀ 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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), t1 = K, value(ks(1) .2), value(t2), ap(ks(1) .1), ap(ks(1)), ks(1) .1 = t1 _ t2 ⊦ 〈RED(t2; t0)〉 t0 = K _ K

Rule argument

[x0lemmaarg: spec-name: term instance-name: lemma-name: elim antecedent: succedent: ap(t) → (t0 = t .1 ∧ t1 = t .2 ↔ t = t0 _ t1) substlist: [t, t0, t1] → [ks(1) .1, t1, t2] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ ap(t) → ( t0 = t .1 ∧ t1 = t .2 ↔ t = t0 _ t1)])], [EXTRATERMS : (ks(1) .1, t1, t2)]]