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Proof step 18 from proof fill-sgn-01 (previous step)

(¬ boolvar) ∈ bits0 → (∃ bits2, bits1. bits0 = bits2 + (¬ boolvar)' + bits1 ∧ ¬ (¬ boolvar) ∈ bits2), n1 + # bits0 = n + # sgn(boolvar ' + bits0), # sgn(boolvar ' + bits0) ≤ n1 + # bits0, # sgn(boolvar ' + bits0) ≤ 1 + # bits0, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ n1 = 0, mklist(boolvar, n) + sgn(boolvar ' + bits0) = mklist(boolvar, n1) + bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n1 + # bits0 = n + # sgn(boolvar ' + bits0), # sgn(boolvar ' + bits0) ≤ n1 + # bits0, # sgn(boolvar ' + bits0) ≤ 1 + # bits0, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0), (¬ boolvar) ∈ bits0 → bits0 = bits + (¬ boolvar)' + bits1 ∧ ¬ (¬ boolvar) ∈ bits ⊦ n1 = 0, mklist(boolvar, n) + sgn(boolvar ' + bits0) = mklist(boolvar, n1) + bits0

Proof step 18 from proof fill-sgn-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information