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Proof step 4 from proof termination-RED1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ t0. #(t0) < #(t1 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t1 .1 = K, ap(t1), value(t1 .2), value(t2), ¬ hasS(t1), ¬ hasS(t2), ¬ (¬ hasS(t1 .2) ∧ ¬ #(t1) + #(t2) < #(t1 .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)])]]