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Proof step 102 from proof ks-RED (previous step)

∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n ≠ 1, n ≠ 0 ⊦ 〈RED1(ks(n); tn0)〉〈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 lemma

And obtained the following new goals

Proof step 93:
⊦ 〈RED1(ks(n); tn)〉 ((tn = none → n ≤ 1) ∧ (tn ≠ none → ¬ n ≤ 1 ∧ tn = some(ks(n - 2))))
Proof step 101:
〈RED1(ks(n); tn)〉 ((tn = none → n ≤ 1) ∧ (tn ≠ none → ¬ n ≤ 1 ∧ tn = some(ks(n - 2)))), ∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n ≠ 1, n ≠ 0 ⊦ 〈RED1(ks(n); tn0)〉〈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: instance-name: lemma-name: ks-RED1 antecedent: succedent: 〈RED1(ks(n); tn)〉 ( (tn = none → n ≤ 1) ∧ (tn ≠ none → ¬ n ≤ 1 ∧ tn = some(ks(n - 2)))) substlist: [n] → [n] add preconditions?: F is rewrite lemma?: F current spec?: T rotate lemma?: F replace all?: F paths: () dummy: () ]

Additional information

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