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Proof step 10 from proof validvals-append (previous step)

∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ (rk × val ∈ rkvs → validval(str, rk, val, st)) ∧ (rkv0 ∈ rkvs0 → validval(str, rkv0 .1, rkv0 .2, st))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
rkv0 = rk0 × val0, ∀ rkv. rkv ∈ rkvs ∨ rkv ∈ rkvs0 → validval(str, rkv .1, rkv .2, st) ⊦ (rk × val ∈ rkvs → validval(str, rk, val, st)) ∧ (rkv0 ∈ rkvs0 → validval(str, rk0, val0, st))

Rule argument

[speclemmaarg: ,rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val ,[rkv, rk, val] → [rkv0, rk0, val0] ,T ,pair ,refkeyvalues ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val])], [EXTRATERMS : (rkv0, rk0, val0)]]