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Proof step 5 from proof validvals-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
rk × val ∈ rkvs, ∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ validval(str, rk, val, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]