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Proof step 54 from proof bitsor-sgn-in (previous step)

# bits0 = # bits1 ⊦ bits2 + bits0 = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits2 + bits0, false ' + bits1), # bits2 = 1, bits2 = [], true ∈ bits2, true ∈ bits0, bits0 ≠ [] ∧ bits1 ≠ []

User applied rule cut formula

And obtained the following new goals

Proof step 55:
# bits0 = # bits1 ⊦ bits0 ≠ [], bits2 + bits0 = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits2 + bits0, false ' + bits1), # bits2 = 1, bits2 = [], true ∈ bits2, true ∈ bits0, bits0 ≠ [] ∧ bits1 ≠ []

Rule argument

fma: bits0 ≠ []

Additional information

Used simplifier rules: ⊦ # [] = 0;% ⊦ # bits = 0 ↔ bits = []; Used formulas: bits0 ≠ []