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Proof step 9 from proof validvals-put-any (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
rkv = rk0 × val0, validval(str, rk, val, st), validvals(str, rkvs, st), rkv ∈ rkvs ⊦ validval(str, rk0, val0, st[rk, val])

Rule argument

[speclemmaarg: ,rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val ,[rkv, rk, val] → [rkv, rk0, val0] ,T ,pair ,refkeyvalues ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val])], [EXTRATERMS : (rkv, rk0, val0)]]