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Proof step 6 from proof sgn-last-diff (previous step)

⊦ sgn(boolvar0 ' + boolvar ') = sgn(boolvar0 ' + bits0) + boolvar ', (¬ boolvar0) ∈ bits0, ¬ (¬ boolvar) ∈ bits0 ∧ ¬(¬ boolvar ↔ boolvar0)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
⊦ ¬ (¬ boolvar) ∈ bits0, sgn(boolvar0 ' + boolvar ') = sgn(boolvar0 ' + bits0) + boolvar ', (¬ boolvar0) ∈ bits0
Proof step 15:
⊦ ¬(¬ boolvar ↔ boolvar0), sgn(boolvar0 ' + boolvar ') = sgn(boolvar0 ' + bits0) + boolvar ', (¬ boolvar0) ∈ bits0

Rule argument

pos: right 3

Additional information