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Proof step 80 from proof termination-RED1 (previous step)

〈RED1(t2; tn)〉 tn = tn2, 〈RED1(t3; tn1)〉 tn1 = none, ∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t3 ≠ K, value(t3 _ t4), ¬ hasS(t2), ¬ value(t3), ¬ hasS(t3), ¬ hasS(t4), ¬ (ap(t3) ∧ t3 .1 = S), ¬ (ap(t3) ∧ t3 .1 = K ∧ value(t3 .2) ∧ value(t4)), ¬ (ap(t3) ∧ ap(t3 .1) ∧ t3 .1 .1 = S), tn2 ≠ none → ¬ hasS(tn2) ∧ #(tn2) < #(t2), t = t3 _ t4, ¬ tn2 ≠ none ⊦ 〈skip〉 (tn2 ≠ none → ¬ hasS(tn2) ∧ ¬ #(t) + #(t2) < #(tn2))

Proof step 80 from proof termination-RED1 (previous step)

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

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