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Proof step 17 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)), ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S, v = t .2, v0 = t .1 .2, v4 = t .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v), tn = some(v4 _ v _ (v0 _ v)) ⊦ 〈skip〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t))

Heuristic symbolic execution applied rule skip right

And obtained the following new goals

Proof step 16:
∀ 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)), ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S, v = t .2, v0 = t .1 .2, v4 = t .1 .1 .2, value(v4) ∧ value(v0) ∧ value(v), tn = some(v4 _ v _ (v0 _ v)), ¬ (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]