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Proof step 70 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) = t1 _ t2, ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S) ⊦ 〈RED1(t1; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(t1) then { RED1(t2; 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))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 35:
⊦ n ≠ 0 → ks(n) = ks(n -1) _ K
Proof step 69:
∀ n0. n0 < 1 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), ks(1 -1) _ K = t1 _ t2, ¬ (ap(t1) ∧ t1 .1 = K ∧ value(t1 .2) ∧ value(t2)), ¬ (ap(t1) ∧ ap(t1 .1) ∧ t1 .1 .1 = S), 1 ≠ 0 ⊦ 〈RED1(t1; tn0)〉〈n := 1 ; if tn0 ≠ none then tn0 := some(tn0 .t _ ks(n) .2) else if value(t1) then { RED1(t2; 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: termfuns instance-name: lemma-name: ks-rec-01 antecedent: succedent: n ≠ 0 → ks(n) = ks(n -1) _ K substlist: [n] → [1] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 1, 1, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (1)]]