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Proof step 6 from proof validvals-append (previous step)

rk0 × val0 ∈ rkvs0, ∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ validval(str, rk0, val0, st)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 2, [QUANTTERMLIST : (rk0 × val0)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rk × val) .1 = rk], [CSIMPSEQ : ⊦ (rk × val) .2 = val])], [EXTRATERMS : (rk0 × val0)]]