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Proof step 12 from proof validvals-put-any (conclusion)

⊦ validval(str, rk, val, st) ∧ validvals(str, rkvs, st) → validvals(str, rkvs, st[rk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
validvals(str, rkvs, st), validval(str, rk, val, st) ⊦ validvals(str, rkvs, st[rk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]