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Proof step 6 from proof at-put-last-other-01 (conclusion)

⊦ (r0 ≠ st[r -- sks .butlast] .refval ∨ sk ≠ sks .last) ∧ 1 < # sks → (st[r -- sks, val][r0 -- sk]) = st[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
r0 ≠ st[r -- sks .butlast] .refval ∨ sk ≠ sks .last ⊦ (st[r -- sks, val][r0 -- sk]) = st[r0 -- sk], # sks = 1, sks = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # sks = 0 ↔ sks = []], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)])]]