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Proof step 13 from proof bitsxor-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
bitsxor(false ', bits0) = [] ⊦ # bits0 = 1, bits0 = false ', bits0 = true ', bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ # boolvar ' = 0 +1;% ⊦ # bits = 1 ∧ bits ≠ true ' → bits = false '; (forward)