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Proof step 2 from proof validvals-append-01 (conclusion)

⊦ validvals(str, rkvs + rkvs0, st) ↔ ¬ ¬ (validvals(str, rkvs, st) ∧ validvals(str, rkvs0, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1])]]