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Proof step 12 from proof put-rkvs-switch (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ val, rk, st, rkvs0. # rkvs0 < # rkvs → (true → (st[rk, val][rkvs0]) = (st[rkvs0][rk, val]) ∨ rk ∈ rkvs0) ⊦ rkvs = [], (st[rk, val][rkvs .butlast][rkvs .last .1, rkvs .last .2]) = (st[rkvs][rk, val]), rk ∈ rkvs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]