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Proof step 12 from proof bitsxor-is-orand (previous step)

¬(bitsinvert(bits) ≠ [] ∧ bits0 = false '), ∀ bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits2int(bitsxor(bits, bits0)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits0), bitsand(bitsinvert(bits0), bits))), bits0 = [], bits = [], bits0 = true ', # bits ≠ 1 ∧ # bits0 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bits = 1, ∀ bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits2int(bitsxor(bits, bits0)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits0), bitsand(bitsinvert(bits0), bits))), bits0 = [], bits = [], 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)% ⊦ bitsinvert(bits) = [] ↔ bits = [];