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Proof step 1 from proof fillsgn-butlast (conclusion)

⊦ 1 < # bytes → fillsgn(n, bytes.butlast) = fillsgn(n + 1, bytes).butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ fillsgn(n, bytes.butlast) = fillsgn(n + 1, bytes).butlast, # bytes = 1, bytes = []

Rule argument

Additional information

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