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Proof step 5 from proof fill-sgn-01 (previous step)

n = n1 + # bits0 ∧ # bits0 ≤ n, # bits0 < n, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ fillsgn(n, sgn(boolvar ' + bits0)) = mklist(boolvar, n1) + bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ fillsgn(n1 + # bits0, sgn(boolvar ' + bits0)) = mklist(boolvar, n1) + bits0, n1 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n → m ≤ n;