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Proof step 6 from proof bitsxor-is-orand (previous step)

¬(1 < # bitsinvert(bits) ∧ 1 < # bits0), ∀ 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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ 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 = [], # bits ≠ 1 ∧ # bits0 ≠ 1

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # bitsinvert(bits) = # bits;