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Proof step 5 from proof TOPRED-lem (previous step)

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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
t .1 .1 = K, ap(t), ap(t .1), value(t .1 .2), value(t .2) ⊦ t =>top t .1 .2

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ some(t) .t = t], [CSIMPSEQ : ⊦ some(t) ≠ none])]]