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Proof step 34 from proof fill-sgn-01 (previous step)

¬ (¬ boolvar) ∈ bits0, n1 + # bits0 = n + # boolvar ', # boolvar ' ≤ n1 + # bits0, # boolvar ' ≤ 1 + # bits0, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ n1 = 0, mklist(boolvar, n) + boolvar ' = mklist(boolvar, n1) + bits0, (¬ boolvar) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
n1 + # bits0 = 1 + n, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ boolvar ' + mklist(boolvar, n) = mklist(boolvar, n1) + bits0, n1 = 0, (¬ boolvar) ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m + n;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ # boolvar ' = 0 +1;