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Proof step 40 from proof or-invandinv (previous step)

bits0 = bits + boolvar0 ', ∀ bits1, bits0. # bits1 < #(bits2 + boolvar ') → true → bits1 = [] ∨ bits0 = [] ∨ bitsor(bits1, bits0) = bitsinvert(bitsand(bitsinvert(bits1), bitsinvert(bits0))) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = bits0.first ', bits0 = [], bitsor(bits2, bits) + (boolvar ∨ boolvar0)' = bitsinvert(bitsand(bitsinvert(bits2) + (¬ boolvar)', bitsinvert(bits0)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
∀ bits1, bits0. # bits1 < #(bits2 + boolvar ') → true → bits1 = [] ∨ bits0 = [] ∨ bitsor(bits1, bits0) = bitsinvert(bitsand(bitsinvert(bits1), bitsinvert(bits0))) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits + boolvar0 ' = (bits + boolvar0 ').first ', bitsor(bits2, bits) + (boolvar ∨ boolvar0)' = bitsinvert(bitsand(bitsinvert(bits2) + (¬ boolvar)', bitsinvert(bits) + (¬ boolvar0)'))

Rule argument

Additional information

Used simplifier rules: ⊦ bitsinvert(boolvar ') = (¬ boolvar)';% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];