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Proof step 35 from proof ks-RED1 (previous step)

〈RED1(ks(1); tn)〉 tn = none, ks(1) .1 = S _ t1, ap(ks(1)), ¬ value(ks(1)), ¬ (value(t1) ∧ value(ks(1) .2)) ⊦

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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), t, t0] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

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