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Proof step 18 from proof bitsxor-is-orand (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ # bits0 = 1, bits0 = false ', bits0 = true ', bits0 = [], bits2int(bitsxor(true ', bits0)) = bits2int(bitsor(false ', bitsinvert(bits0)))

Rule argument

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = [];