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Proof step 22 from proof sgn-last-04 (previous step)

bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(boolvar1 ' + bits) = sgn(boolvar1 ' + bits0) + boolvar0 ', bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0, ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1)

User applied rule cut formula

And obtained the following new goals

Proof step 23:
bits0 = [], bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ sgn(boolvar1 ' + bits) = sgn(boolvar1 ' + bits0) + boolvar0 ', bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0, ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1)
Proof step 25:
bits0 + boolvar0 ' = boolvar1 ' + bits, ∀ boolvar. (¬ boolvar) ∈ bits0 → sgn(bits0 + boolvar ') = sgn(bits0) + boolvar ' ⊦ bits0 = [], sgn(boolvar1 ' + bits) = sgn(boolvar1 ' + bits0) + boolvar0 ', bits0 ≠ [] ∧ (bits0.first ↔ ¬ boolvar1), ¬(boolvar1 ↔ boolvar0) ∧ ¬ boolvar1 ∈ bits0, ¬ boolvar0 ∈ bits ∧ ¬(boolvar0 ↔ boolvar1), ¬ (¬ boolvar0) ∈ bits0 ∧ ¬(¬ boolvar0 ↔ boolvar1)

Rule argument

fma: bits0 = []

Additional information

Used formulas: bits0 = []