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Proof step 10 from proof bitsxor-empty (previous step)

¬(bits ≠ [] ∧ bits0 = false '), bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = [], bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
bitsxor(bits, bits0) = [], # bits = 1 ⊦ bits = [], bits0 = [], bits0 = true ', bits0 = false ', # bits0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 1 ∧ bits ≠ true ' → bits = false '; (forward)% ⊦ # bits = 1 ∧ bits ≠ false ' → bits = true '; (forward)