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Proof step 15 from proof in-put-rkvs (previous step)

∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk ∈ (st[rkv .1, rkv .2][rkvs0]) ↔ rk ∈ st ∨ rk ∈ rkv ' ∨ rk ∈ rkvs0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
rkv = rk0 × val, ∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk ∈ (st[rk0, val][rkvs0]) ↔ rk ∈ st ∨ rk ∈ rkv ' ∨ rk ∈ rkvs0

Rule argument

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

Additional information

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