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

¬ hasS(t), ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2) ⊦ 〈tn0 := some(t .1 .2)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 5:
¬ hasS(t), ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2), ¬ (some(t .1 .2) ≠ none → ¬ hasS(some(t .1 .2)) ∧ #(some(t .1 .2)) < #(t)) ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]