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Proof step 32 from proof termination-RED1 (previous step)

〈RED1(t1; tn)〉 tn = tn0, ∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), tn0 ≠ none, #(tn0) < #(t1), ¬ hasS(tn0), ¬ hasS(t2), ¬ hasS(t1), ¬ (¬ hasS(tn0 .t) ∧ #(tn0 .t) < #(t1)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S), ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)) ⊦

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[x0lemmaarg: spec-name: term_or_none instance-name: lemma-name: elim antecedent: succedent: tn ≠ none → (t = tn .t ↔ tn = some(t)) substlist: [tn, t] → [tn0, t] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ some(t) ≠ none], [CSIMPSEQ : ⊦ #(some(t)) = #(t)], [CSIMPSEQ : ⊦ hasS(some(t)) ↔ hasS(t)], [CSIMPELIM : ⊦ tn ≠ none → (t = tn .t ↔ tn = some(t))])], [EXTRATERMS : (tn0, t)]]