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Proof step 96 from proof ks-RED (previous step)

〈RED1(ks(n); tn)〉 tn = some(ks(n1)), ∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n ≠ 0, n ≠ 1, n = n1 + 2, 2 ≤ n ⊦ 〈RED(ks(n1); t0)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 95:
〈RED1(ks(n1 + 2); tn)〉 tn = some(ks(n1)), ∀ n0. n0 < n1 + 2 → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))) ⊦ 〈RED(ks(n1); t0)〉 ((even(n1) → t0 = K) ∧ (odd(n1) → t0 = K _ K))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ odd(m + 2) ↔ odd(m)], [CSIMPSEQ : ⊦ even(m + 2) ↔ even(m)], [CSIMPSEQ : ⊦ m ≤ m + n], [CSIMPSEQ : ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : n < n0 ⊦ ((* n)) ≠ m0 + ((* n0))])]]