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Proof step 22 from proof fill-last (previous step)

bits = boolvar0 + bits0, # bits ≤ m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits = [], fillsgn(2 + m0, bits + boolvar ') = boolvar0 ' + fillsgn(m0, bits) + boolvar '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
# bits0 < m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ fillsgn(2 + m0, (boolvar0 ' + bits0) + boolvar ') = boolvar0 ' + fillsgn(m0, boolvar0 ' + bits0) + boolvar '

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;