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Proof step 8 from proof validvals-put-any (previous step)

rkv = rk0 × val0, validval(str, rk, val, st), validvals(str, rkvs, st), rkv ∈ rkvs ⊦ validval(str, rk0, val0, st[rk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
validval(str, rk, val, st), validvals(str, rkvs, st), rk0 × val0 ∈ rkvs ⊦ validval(str, rk0, val0, st[rk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]