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Proof step 29 from proof ks-RED1 (previous step)

〈RED1(ks(n0); tn)〉 tn = tn1, ks(n0) .1 .1 = S, ap(ks(n0)), ap(ks(n0) .1), tn0 = none → n0 ≤ 1, tn0 ≠ none → ¬ n0 ≤ 1 ∧ tn0 = some(ks(n0 - 2)), ¬ (ks(n0) .1 = K ∧ value(ks(n0) .2)), ¬ (value(ks(n0) .1 .2) ∧ value(ks(n0) .2)), tn1 = tn0, tn1 ≠ none, ¬ (some(tn1 .t _ K) = none → n0 = 0) ∨ ¬ (some(tn1 .t _ K) ≠ none → n0 ≠ 0 ∧ some(tn1 .t _ K) = some(ks(n0 + 1 - 2))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
〈RED1(ks(n0); tn)〉 tn = some(ks(n0 - 2)), ks(n0) .1 .1 = S, n0 ≠ 0, n0 ≠ 1, ks(n0 - 2) _ K ≠ ks(n0 + 1 - 2), ap(ks(n0) .1), ap(ks(n0)), ¬ (value(ks(n0) .1 .2) ∧ value(ks(n0) .2)), ¬ (ks(n0) .1 = K ∧ value(ks(n0) .2)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ some(t) .t = t], [CSIMPSEQ : ⊦ some(t) = some(t0) ↔ t = t0], [CSIMPSEQ : ⊦ some(t) ≠ none], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)], [CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m])]]