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Proof step 16 from proof validvals-append (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: left 3, ()]

Additional information

[[CONCRETESIMPRULES : ()]]