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Proof step 16 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), ¬ hasS(t), ¬ (ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)), ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S, v = t .2, v0 = t .1 .2, v4 = t .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v), tn = some(v4 _ v _ (v0 _ v)), ¬ (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t .1 .1 .1 = S, t .1 .1 ≠ K, ap(t), ap(t .1), ap(t .1 .1), value(t .1 .1 .2), value(t .1 .2), value(t .2), ¬ hasS(t), ¬ (¬ hasS(t .1 .1 .2 _ t .2 _ (t .1 .2 _ t .2)) ∧ 3 + #(t .1 .1 .2) + #(t .1 .2) + 2 * #(t .2) < #(t)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + m = 2 * m], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ #(t _ t0) = (#(t) + #(t0)) +1], [CSIMPSEQ : ⊦ #(some(t)) = #(t)], [CSIMPSEQ : ⊦ hasS(some(t)) ↔ hasS(t)], [CSIMPSEQ : ⊦ some(t) ≠ none])]]