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Proof step 9 from proof validvals-append (previous step)

rkv0 = rk0 × val0, ∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ (rk × val ∈ rkvs → validval(str, rk, val, st)) ∧ (rkv0 ∈ rkvs0 → validval(str, rk0, val0, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ (rk × val ∈ rkvs → validval(str, rk, val, st)) ∧ (rk0 × val0 ∈ rkvs0 → validval(str, rk0, val0, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]