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Proof step 6 from proof termination-RED1 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t .1 .1 = K, ap(t), ap(t .1), value(t .1 .2), value(t .2), ¬ hasS(t), ¬ (¬ hasS(t .1 .2) ∧ #(t .1 .2) < #(t)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ #(some(t)) = #(t)], [CSIMPSEQ : ⊦ hasS(some(t)) ↔ hasS(t)], [CSIMPSEQ : ⊦ some(t) ≠ none])]]