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Proof step 10 from proof bitsor-sgn-in (previous step)

⊦ bits = [], bits0 = [], boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0), true ∈ bits ∧ false ∈ bits

User applied rule cut formula

And obtained the following new goals

Proof step 11:
true ∈ bits ⊦ bits = [], bits0 = [], boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0), true ∈ bits ∧ false ∈ bits
Proof step 36:
⊦ true ∈ bits, bits = [], bits0 = [], boolvar ∈ bitsor(sgn(bits), bits0) ↔ boolvar ∈ bitsor(bits, bits0), true ∈ bits ∧ false ∈ bits

Rule argument

fma: true ∈ bits

Additional information

Used formulas: true ∈ bits