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Proof step 33 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(false ', 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 34:
∀ bits, bits1. # bits1 < # false ' → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits = [], bits2int(bitsxor(bits, false ')) = bits2int(bitsor(false ', bits))

Rule argument

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;% ⊦ # boolvar ' = 0 +1;% ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ' ≠ [];% ⊦ bitsinvert(false ') = true ';