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Proof step 5 from proof validvals-put-any (previous step)

validval(str, rk, val, st), rk0 × val0 ∈ rkvs → validval(str, (rk0 × val0) .1, (rk0 × val0) .2, st), rk0 × val0 ∈ rkvs ⊦ validval(str, rk0, val0, st[rk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
validval(str, rk, val, st), rk0 × val0 ∈ rkvs, validval(str, rk0, val0, st) ⊦ validval(str, rk0, val0, st[rk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rk × val) .2 = val], [CSIMPSEQ : ⊦ (rk × val) .1 = rk])]]