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Proof step 51 from proof kappa-TOPRED (previous step)

〈let v1 = t .1 .1 .2, v2 = t .1 .2, v3 = t .2 in if value(v1) ∧ value(v2) ∧ value(v3) then tn := some(v1 _ v3 _ (v2 _ v3)) else tn := none〉 tn = tn0, tn0 ≠ none, n ≠ 0, ¬ (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), ¬ (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 ⊦

Heuristic symbolic execution applied rule vardecls left

And obtained the following new goals

Proof step 50:
〈if value(v4) ∧ value(v0) ∧ value(v) then tn := some(v4 _ v _ (v0 _ v)) else tn := none〉 tn = tn0, tn0 ≠ none, n ≠ 0, ¬ (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), ¬ (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 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]