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Proof step 12 from proof termination-RED1 (previous step)

∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t3 .1 = S, t3 ≠ K, ap(t1), ap(t3), value(t3 .2), value(t4), value(t2), ¬ hasS(t1), ¬ hasS(t2), ¬ (¬ hasS(t3 .2 _ t2 _ (t4 _ t2)) ∧ ¬ #(t1) ≤ #(t2) + #(t3 .2) + #(t4) + 2), t1 = t3 _ t4 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t3 .1 = S, t3 ≠ K, ap(t3), value(t3 .2), value(t4), value(t2), ¬ hasS(t2), ¬ hasS(t3), ¬ hasS(t4), ¬ (¬ hasS(t3 .2 _ t2 _ (t4 _ t2)) ∧ ¬ #(t3) ≤ #(t2) + #(t3 .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)])]]