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

ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2) ⊦ 〈tn0 := some(t .1 .2)〉 (tn0 = none ⊃ ¬ (∃ t0. t =>top t0); t =>top tn0 .t)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 5:
ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2), some(t .1 .2) = none ⊃ (∃ t0. t =>top t0) ; ¬ t =>top some(t .1 .2) .t ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]