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Proof step 19 from proof validvals-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]