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Proof step 24 from proof in-put (conclusion)

⊦ 1 < # sks → (rk ∈ st[r -- sks, val] ↔ rk ∈ st ∨ rk = st[r -- sks .butlast] .refval -- sks .last)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ rk ∈ st[r -- sks, val] ↔ rk ∈ st ∨ rk = st[r -- sks .butlast] .refval -- sks .last, # sks = 1, sks = []

Rule argument

[emptyarg]

Additional information

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