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Proof step 24 from proof ks-RED (previous step)

∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), ks(1) .1 = S _ t2 _ t3, ap(ks(1)), ¬ (value(t2) ∧ value(t3) ∧ value(ks(1) .2)) ⊦ 〈RED1(S _ t2 _ t3; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(S _ t2 _ t3) then { RED1(ks(1) .2; tn0) ; if tn0 ≠ none then tn0 := some(ks(n) .1 _ tn0 .t) }〉〈if tn0 ≠ none then RED(tn0 .t; t0) else t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

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] → [ks(1), t1, t4] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ks(n) ≠ S _ t1 _ t2 _ t3], [CSIMPELIM : ⊦ ap(t) → ( t0 = t .1 ∧ t1 = t .2 ↔ t = t0 _ t1)])], [EXTRATERMS : (ks(1), t1, t4)]]