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Proof step 22 from proof RED-correct (previous step)

〈RED(tn .t; t0): kappa0 〉 t0 = t1, 〈RED1(t; tn0)〉 tn0 = tn, 〈RED1(t; tn0): kappa0 〉 tn0 = tn, ∀ t0, t2. ¬ 〈RED(t0; t'): kappa0 〉 t' = t2 ∨ t0 =>* t2 ∧ irred(t2), tn ≠ none, t => tn .t, ¬ t =>* t1 ∨ ¬ irred(t1) ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
〈RED(t3; t0): kappa0 〉 t0 = t1, 〈RED1(t; tn0)〉 tn0 = tn, 〈RED1(t; tn0): kappa0 〉 tn0 = tn, ∀ t0, t2. ¬ 〈RED(t0; t'): kappa0 〉 t' = t2 ∨ t0 =>* t2 ∧ irred(t2), tn ≠ none, t => t3, ¬ t =>* t1 ∨ ¬ irred(t1), tn = some(t3) ⊦

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] → [tn, t3] 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 : (tn, t3)]]