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Proof step 26 from proof not-value-02 (previous step)

value(C0[S _ t _ t0 _ t1]), ∀ t, t0, t1. ¬ value(C0[S _ t _ t0 _ t1]) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [QUANTTERMLIST : (t, t0, t1)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (t, t0, t1)]]