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Proof step 10 from proof put-rkvs-switch (previous step)

rkvs .last = rk0 × val0, ∀ val, rk, st, rkvs0. # rkvs0 < # rkvs → (true → (st[rk, val][rkvs0]) = (st[rkvs0][rk, val]) ∨ rk ∈ rkvs0) ⊦ rkvs = [], (st[rk, val][rkvs .butlast][rk0, val0]) = (st[rkvs][rk, val]), rk ∈ rkvs

User applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
rkvs = rkvs1 + rkv ', rkv = rk0 × val0, ∀ val, rk, st, rkvs0. # rkvs0 < # rkvs → (true → (st[rk, val][rkvs0]) = (st[rkvs0][rk, val]) ∨ rk ∈ rkvs0) ⊦ rkvs = [], (st[rk, val][rkvs1][rk0, val0]) = (st[rkvs][rk, val]), rk ∈ 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)]]