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Proof step 17 from proof termination-TOPRED (
previous step
)
¬ 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) ⊦ 〈tn0 := none〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ #(tn0) < #(t))
Heuristic symbolic execution applied rule assign right
And closed the goal!
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ()]]