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Proof step 13 from proof put-rkvs-switch (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ rkvs ≠ [] → st[rkvs] = (st[rkvs .butlast][rkvs .last .1, rkvs .last .2])
Proof step 2:
⊦ st[[]] = st
Proof step 12:
rkvs ≠ [], ∀ val, rk, st, rkvs0. # rkvs0 < # rkvs → (true → (st[rk, val][rkvs0]) = (st[rkvs0][rk, val]) ∨ rk ∈ rkvs0) ⊦ (st[rk, val][rkvs .butlast][rkvs .last .1, rkvs .last .2]) = (st[rkvs][rk, val]), rk ∈ rkvs

Rule argument

[crewrite-arg: put-rkvs-last, F, [rkvs, st] → [rkvs, st[rk, val]]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rkvs, st[rk, val])]]