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Proof step 4 from proof validvals-one (conclusion)

⊦ validvals(str, rkv ', st) ↔ validval(str, rkv .1, rkv .2, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
rkv = rk × val ⊦ validvals(str, rkv ', st) ↔ validval(str, rk, val, st)

Rule argument

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

Additional information

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