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Proof step 28 from proof bitsor-sgn-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
# bits0 = # bits1, true ∈ bits2 ⊦ bits2 + bits0 = [], boolvar ↔ boolvar ∈ bitsor(bits2 + bits0, true ' + bits1), # bits2 = 1, bits2 = [], false ∈ bits2, false ∈ bits0, bits0 ≠ [] ∧ bits1 ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];