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

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]

User applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
rkvs = rkvs1 + rkv ', rkv = rk0 × val0, rk ∈ rkvs, ∀ val, st, rk, rkvs0. # rkvs0 < # rkvs → (rk ∈ rkvs0 → (st[rk, val][rkvs0]) = st[rkvs0]) ⊦ rkvs = [], (st[rk, val][rkvs1][rk0, val0]) = st[rkvs]

Rule argument

[speclemmaarg: rkvs ≠ [] ,rkv = rkvs .last ∧ rkvs0 = rkvs .butlast ↔ rkvs = rkvs0 + rkv ' ,[rkvs, rkv, rkvs0] → [rkvs, rkv, rkvs1] ,T ,list-last ,refkeyvalues-assoclista ,last-elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : rkvs ≠ [] ⊦ rkv = rkvs .last ∧ rkvs0 = rkvs .butlast ↔ rkvs = rkvs0 + rkv '])], [EXTRATERMS : (rkvs, rkv, rkvs1)]]