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Proof step 12 from proof termination-TOPRED (previous step)

¬ 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), ¬ (some(v4 _ v _ (v0 _ v)) ≠ none → ¬ hasS(some(v4 _ v _ (v0 _ v))) ∧ #(some(v4 _ v _ (v0 _ v))) < #(t)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
t .1 .1 .1 = S, ap(t), ap(t .1), ap(t .1 .1), value(t .1 .1 .2), value(t .1 .2), value(t .2) ⊦ t .1 .1 = K, 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])]]