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Proof step 21 from proof bitsxor-is-orand (previous step)

bitsinvert(bits0) ≠ [] ∧ false ' = false ', ∀ 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(bitsinvert(bits0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
∀ 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 = [], ~1 = bits2int(bits0) + bits2int(bitsxor(true ', bits0))

Rule argument

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ ~ i - j = ~(i + j);% ⊦ bits ≠ [] → bits2int(bitsinvert(bits)) = ~ bits2int(bits) - 1;