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Proof step 3 from proof termination-TOPRED (previous step)

t = t0 _ t1, t0 .1 = K, ap(t), ap(t0), value(t0 .2), value(t1) ⊦ hasS(t), ¬ hasS(t0 .2) ∧ #(t0 .2) < #(t)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
t0 .1 = K, ap(t0), value(t0 .2), value(t1) ⊦ hasS(t0), hasS(t1), ¬ hasS(t0 .2) ∧ ¬ #(t0) + #(t1) < #(t0 .2)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ #(t _ t0) = (#(t) + #(t0)) +1], [CSIMPSEQ : ⊦ hasS(t _ t0) ↔ hasS(t) ∨ hasS(t0)], [CSIMPSEQ : ⊦ ap(t _ t0)])]]