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Proof step 2 from proof bnotempty (previous step)

x.butlast = [], 1 < # x ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x.butlast = [] ⊦ # x = 1, x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)