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Proof step 26 from proof fill-last (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: boolvar0 ' + bits0, boolvar

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ []; Used terms: boolvar0 ' + bits0, boolvar