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Proof step 6 from proof termination-RED (previous step)

〈RED1(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), tn1 ≠ none, #(tn1) < #(t), ¬ hasS(tn1), ¬ hasS(t) ⊦ 〈RED(tn1 .t; t0)〉 true

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
〈RED1(t; tn)〉 tn = tn1, ∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), tn1 ≠ none, #(tn1) < #(t), ¬ hasS(tn1), ¬ hasS(t), tn1 = some(t1) ⊦ 〈RED(t1; t0)〉 true

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] → [tn1, t1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ tn ≠ none → (t = tn .t ↔ tn = some(t))])], [EXTRATERMS : (tn1, t1)]]