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Proof step 14 from proof fill-last (previous step)

¬(bits ≠ [] ∧ # bits < m0 + 1), ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = fillsgn(m0 + 1, bits) + boolvar '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m0 < # bits, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ fillsgn(2 + m0, bits + boolvar ') = fillsgn(m0 + 1, bits) + boolvar ', bits = []

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n + 1 ↔ ¬ n < m;