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Proof step 12 from proof validvals-putrkvs (conclusion)

⊦ validvals(str, rkvs0, st) ∧ validvals(str, rkvs1, st) → validvals(str, rkvs0, st[rkvs1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
validvals(str, rkvs1, st), validvals(str, rkvs0, st) ⊦ validvals(str, rkvs0, st[rkvs1])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]