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Proof step 11 from proof fill-sgn-01 (previous step)

# sgn(boolvar ' + bits0) ≤ n1 + # bits0 ∧ sgn(boolvar ' + bits0) ≠ [], ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ mklist(sgn(boolvar ' + bits0).first, n1 + # bits0 - # sgn(boolvar ' + bits0)) + sgn(boolvar ' + bits0) = mklist(boolvar, n1) + bits0, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# sgn(boolvar ' + bits0) ≤ n1 + # bits0, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ n1 = 0, mklist(boolvar, n1 + # bits0 - # sgn(boolvar ' + bits0)) + sgn(boolvar ' + bits0) = mklist(boolvar, n1) + bits0

Rule argument

Additional information

Used simplifier rules: sgn-first: ⊦ sgn(boolvar ' + bits).first ↔ boolvar;%sgn-not-empty-01: ⊦ sgn(boolvar ' + bits) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;