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Proof step 5 from proof vals-in (previous step)

val0 ∈ fis0 .vals ↔ (∃ sk. sk × val0 ∈ fis0), val ∈ fis0 .vals, sk × val ∈ fis0, ∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ ∃ sk. sk × val = sk0 × val0 ∨ sk × val ∈ fis0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 1, [QUANTTERMLIST : (sk)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk)]]