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Proof step 11 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t3 .1 = S, t3 ≠ K, ap(t3), value(t3 .2), value(t4), value(t2), ¬ hasS(t2), ¬ hasS(t3), ¬ hasS(t4), ¬ (¬ hasS(t3 .2 _ t2 _ (t4 _ t2)) ∧ ¬ #(t3) ≤ #(t2) + #(t3 .2) + 1) ⊦

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] → [t3, t, t1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ t _ t0 ≠ K], [CSIMPSEQ : ⊦ ap(t _ t0)], [CSIMPSEQ : ⊦ hasS(t _ t0) ↔ hasS(t) ∨ hasS(t0)], [CSIMPSEQ : ⊦ hasS(S)], [CSIMPELIM : ⊦ ap(t) → ( t0 = t .1 ∧ t1 = t .2 ↔ t = t0 _ t1)])], [EXTRATERMS : (t3, t, t1)]]