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Proof step 3 from proof bitsxor-append-same-size (previous step)

# bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ bits = [] ∨ bits0 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits0, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits0, bits2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ bitsxor(bits + bits0, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits0, bits2), bits1 = [], bits0 = [], bits = []

Proof step 3 from proof bitsxor-append-same-size (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information