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Proof step 28 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 .1 .1 = S, ap(ks(1) .1 .1), ap(ks(1) .1), ap(ks(1)), ¬ (ks(1) .1 .1 = K ∧ value(ks(1) .1 .2) ∧ value(ks(1) .2)), ¬ (value(ks(1) .1 .1 .2) ∧ value(ks(1) .1 .2) ∧ value(ks(1) .2)) ⊦ 〈RED1(ks(1) .1; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(ks(1) .1) 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 obtained the following new goals

Proof step 27:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), t1 = S, ap(ks(1) .1 .1), ap(ks(1) .1), ap(ks(1)), ¬ (ks(1) .1 .1 = K ∧ value(ks(1) .1 .2) ∧ value(ks(1) .2)), ¬ (value(t2) ∧ value(ks(1) .1 .2) ∧ value(ks(1) .2)), ks(1) .1 .1 = t1 _ t2 ⊦ 〈RED1(ks(1) .1; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(ks(1) .1) 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))

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) .1 .1, 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 : (ks(1) .1 .1, t1, t2)]]