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Proof step 10 from proof validvals-put-any (previous step)

validvals(str, rkvs, st), validval(str, rk, val, st) ⊦ ∀ rkv. rkv ∈ rkvs → validval(str, rkv .1, rkv .2, st[rk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
validval(str, rk, val, st), validvals(str, rkvs, st), rkv ∈ rkvs ⊦ validval(str, rkv .1, rkv .2, st[rk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]