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Proof step 5 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t .1 .1 = K, ap(t), ap(t .1), value(t .1 .2), value(t .2), ¬ hasS(t), ¬ (¬ hasS(t .1 .2) ∧ #(t .1 .2) < #(t)) ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t1 .1 = K, ap(t), ap(t1), value(t1 .2), value(t2), ¬ hasS(t), ¬ (¬ hasS(t1 .2) ∧ #(t1 .2) < #(t)), t = t1 _ t2 ⊦

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] → [t, 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 : (t, t1, t2)]]