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Proof step 8 from proof termination-TOPRED (previous step)

t0 = t2 _ t3, t2 .1 = S, ap(t0), ap(t2), value(t2 .2), value(t3), value(t1) ⊦ t2 = K, hasS(t0), hasS(t1), ¬ hasS(t2 .2 _ t1 _ (t3 _ t1)) ∧ ¬ #(t0) ≤ #(t1) + #(t2 .2) + #(t3) + 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
t2 .1 = S, ap(t2), value(t2 .2), value(t3), value(t1) ⊦ t2 = K, hasS(t1), hasS(t2), hasS(t3), ¬ hasS(t2 .2 _ t1 _ (t3 _ t1)) ∧ ¬ #(t2) ≤ #(t1) + #(t2 .2) + 1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n ≤ m + n0 ↔ n ≤ n0], [CSIMPSEQ : n < n0 ⊦ m + ((* n)) ≤ m0 + ((* n0)) ↔ m ≤ m0 + (n0 - n)], [CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPSEQ : ⊦ #(t _ t0) = (#(t) + #(t0)) +1], [CSIMPSEQ : ⊦ hasS(t _ t0) ↔ hasS(t) ∨ hasS(t0)], [CSIMPSEQ : ⊦ ap(t _ t0)])]]