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Proof step 20 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 ') = (bits.first ' + fillsgn((m0 + 1)-1, bits)) + boolvar '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# bits ≤ m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = bits.first ' + fillsgn(m0, bits) + boolvar '

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (m + 1)-1 = m;