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Proof step 26 from proof bitsor-first-02 (previous step)

bitsor(false ' + bits3, false ' + bits0) = (true ' + bits1).butlast, boolvar ∨ boolvar0 ↔ (true ' + bits1).last, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits0) ∧ ¬ boolvar ∧ ¬ boolvar0

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 27:
bits1 = [], bitsor(false ' + bits3, false ' + bits0) = (true ' + bits1).butlast, boolvar ∨ boolvar0 ↔ (true ' + bits1).last, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits0) ∧ ¬ boolvar ∧ ¬ boolvar0
Proof step 30:
bitsor(false ' + bits3, false ' + bits0) = (true ' + bits1).butlast, boolvar ∨ boolvar0 ↔ (true ' + bits1).last, ∀ bits1, bits2, bits. # bits < #(bits3 + boolvar ') → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ bits1 = [], ¬ boolvar ∧ ¬ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ bitsor(false ' + bits3, false ' + bits0) ∧ ¬ boolvar ∧ ¬ boolvar0

Rule argument

fma: bits1 = []

Additional information

Used formulas: bits1 = []