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Proof step 13 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ (rkv ∈ rkvs → validval(str, rkv .1, rkv .2, st)) ∧ (rkv0 ∈ rkvs0 → validval(str, rkv0 .1, rkv0 .2, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]