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Proof step 4 from proof put-rkvs-last (previous step)

⊦ rkvs = [], st[rkvs] = (st[rkvs .butlast][rkvs .last .1, rkvs .last .2])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
rkvs .last = rk × val ⊦ rkvs = [], st[rkvs] = (st[rkvs .butlast][rk, val])

Rule argument

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

Additional information

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