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Proof step 16 from proof put-rkvs-absorb (previous step)

rk ∈ rkvs, ∀ val, st, rk, rkvs0. # rkvs0 < # rkvs → (rk ∈ rkvs0 → (st[rk, val][rkvs0]) = st[rkvs0]) ⊦ rkvs = [], (st[rk, val][rkvs .butlast][rkvs .last .1, rkvs .last .2]) = st[rkvs]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
rkvs .last = rk0 × val0, rk ∈ rkvs, ∀ val, st, rk, rkvs0. # rkvs0 < # rkvs → (rk ∈ rkvs0 → (st[rk, val][rkvs0]) = st[rkvs0]) ⊦ rkvs = [], (st[rk, val][rkvs .butlast][rk0, val0]) = st[rkvs]

Rule argument

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

Additional information

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