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Proof step 58 from proof bitsor-sgn-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
# bits0 = # bits1 ⊦ # bits2 = 1, bits1 = [], bits0 = [], bits2 = [], bits2 + bits0 = [], ¬ boolvar ∨ true ∈ bits1 ↔ boolvar ∈ bitsor(bits2, false ') ∨ boolvar ∈ bits1, true ∈ bits2, true ∈ bits0

Rule argument

Additional information

Used simplifier rules: bitsor-zeroes-samelen: ⊦ ¬ true ∈ bits ∧ # bits = # bits0 ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsor(bits, bits0) = bits0;% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);