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Proof step 23 from proof validvals-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ (∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st)) ↔ validvals(str, rkvs, st) ∧ validvals(str, rkvs0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1], [CSIMPSEQ : ⊦ rkv ∈ rkvs + rkvs0 ↔ ¬ ¬ (rkv ∈ rkvs ∨ rkv ∈ rkvs0)])]]