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Proof step 7 from proof bitsxor-empty (
previous step
)
¬(bits ≠ [] ∧ bits0 = true '), bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = [], # bits ≠ 1 ∧ # bits0 ≠ 1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
bitsxor(bits, bits0) = [] ⊦ bits = [], bits0 = [], bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1
Rule argument
Additional information