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Proof step 20 from proof validvals-append (previous step)

⊦ (∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st)) ↔ (∀ rkv. rkv ∈ rkvs → validval(str, rkv .1, rkv .2, st)) ∧ (∀ rkv. rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st))

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]