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Proof step 29 from proof bitsor-sgn-in (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 30:
# bits0 = # bits1, true ∈ bits2 ⊦ bits0 ≠ [], bits2 + bits0 = [], boolvar ↔ boolvar ∈ bitsor(bits2 + bits0, true ' + bits1), # bits2 = 1, bits2 = [], false ∈ bits2, false ∈ bits0, bits0 ≠ [] ∧ bits1 ≠ []

Rule argument

fma: bits0 ≠ []

Additional information

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