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Proof step 22 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 tn = tn0, ks(n0) .1 .1 = S, ks(n0) .1 ≠ K, ks(n0) .1 .2 _ K _ (ks(n0) .2 _ K) ≠ ks(n0 + 1 - 2), n0 ≠ 0, ap(ks(n0)), ap(ks(n0) .1), value(ks(n0) .1 .2), value(ks(n0) .2), tn0 = none → n0 = 1, tn0 ≠ none → n0 ≠ 1 ∧ tn0 = some(ks(n0 - 2)) ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
〈RED1(ks(n0); tn)〉 tn = tn0, t0 = S, ks(n0) .1 ≠ K, t1 _ K _ (ks(n0) .2 _ K) ≠ ks(n0 + 1 - 2), n0 ≠ 0, ap(ks(n0)), ap(ks(n0) .1), value(t1), value(ks(n0) .2), tn0 = none → n0 = 1, tn0 ≠ none → n0 ≠ 1 ∧ tn0 = some(ks(n0 - 2)), ks(n0) .1 = t0 _ t1 ⊦

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(n0) .1, t, t0] 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(n0) .1, t, t0)]]