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Proof step 2 from proof bitsor-app (previous step)

true, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ # bits0 = # bits2 ∧ bits ≠ [] ∧ bits0 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits0, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits0, bits2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → true → # bits3 = # bits2 ∧ bits ≠ [] ∧ bits3 ≠ [] ∧ bits1 ≠ [] ∧ bits2 ≠ [] → bitsor(bits + bits3, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits3, bits2) ⊦ bits = [], bits0 = [], bits1 = [], bits2 = [], bitsor(bits + bits0, bits1 + bits2) = bitsor(bits, bits1) + bitsor(bits0, bits2)

Proof step 2 from proof bitsor-app (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information