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Proof step 100 from proof ks-RED (previous step)

〈RED1(ks(n); tn)〉 tn = tn1, ∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n ≠ 1, n ≠ 0, tn1 = none → n ≤ 1, tn1 ≠ none → ¬ n ≤ 1 ∧ tn1 = some(ks(n - 2)) ⊦ 〈if tn1 ≠ none then RED(tn1 .t; t0) else t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 98:
〈RED1(ks(n); tn)〉 tn = tn1, ∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n ≠ 1, n ≠ 0, tn1 = none → n ≤ 1, tn1 ≠ none → ¬ n ≤ 1 ∧ tn1 = some(ks(n - 2)), tn1 ≠ none ⊦ 〈RED(tn1 .t; t0)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))
Proof step 99:
〈RED1(ks(n); tn)〉 tn = tn1, ∀ n0. n0 < n → (true → 〈RED(ks(n0); t)〉 ((even(n0) → t = K) ∧ (odd(n0) → t = K _ K))), n ≠ 1, n ≠ 0, tn1 = none → n ≤ 1, tn1 ≠ none → ¬ n ≤ 1 ∧ tn1 = some(ks(n - 2)), ¬ tn1 ≠ none ⊦ 〈t0 := ks(n)〉 ((even(n) → t0 = K) ∧ (odd(n) → t0 = K _ K))

Rule argument

[fmapos-arg: pos: right 1]

Additional information

[[CONCRETESIMPRULES : ()]]