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Proof step 6 from proof validvals-putrkvs (previous step)

¬ (validval(str, rk, val, st) ∧ validvals(str, rkvs1, st)), validvals(str, rkvs0, st), validvals(str, rkvs1, st), rk × val ∈ rkvs0 ⊦ validval(str, rk, val, st[rkvs1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
validvals(str, rkvs0, st), validvals(str, rkvs1, st), rk × val ∈ rkvs0 ⊦ validval(str, rk, val, st[rkvs1]), validval(str, rk, val, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]