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Proof step 15 from proof bitsxor-is-orand (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
∀ 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 ', bitsand(bitsinvert(bits0), true ')))

Rule argument

Additional information

Used simplifier rules: and-zero-02: ⊦ bits ≠ [] → bitsand(false ', bits) = false ';% ⊦ bitsinvert(true ') = false ';% ⊦ boolvar ' ≠ [];% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ # boolvar ' = 0 +1;% ⊦ # bits = 1 ∧ bits ≠ false ' → bits = true '; (forward)% ⊦ bitsinvert(bits) = true ' ↔ bits = false ';