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Proof step 1 from proof fill-last (conclusion)

⊦ n ≠ 0 ∧ bits ≠ [] → fillsgn(n, bits + boolvar ') = fillsgn(n -1, bits) + boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bits = [], n = 0, fillsgn(n, bits + boolvar ') = fillsgn(n - 1, bits) + boolvar '

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';% ⊦ n -1 = n - 1;