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Proof step 39 from proof bitsxor-is-orand (previous step)

bitsinvert(bits) ≠ [] ∧ bits0 = true ', ∀ 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(bitsinvert(bits), bitsand(bitsinvert(bits0), bits))), bits0 = [], bits = [], # bits ≠ 1 ∧ # bits0 ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
∀ bits, bits1. # bits1 < # true ' → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits = [], bits2int(bitsxor(bits, true ')) = bits2int(bitsor(bitsinvert(bits), false '))

Rule argument

Additional information

Used simplifier rules: and-zero-02: ⊦ bits ≠ [] → bitsand(false ', bits) = false ';% ⊦ # boolvar ' = 0 +1;% ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ boolvar ' ≠ [];% ⊦ bitsinvert(true ') = false ';