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Proof step 43 from proof or-invandinv (previous step)

1 < #(bitsinvert(bits2) + (¬ boolvar)') ∧ 1 < #(bitsinvert(bits) + (¬ boolvar0)'), ∀ 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)').butlast, (bitsinvert(bits) + (¬ boolvar0)').butlast) + ((bitsinvert(bits2) + (¬ boolvar)').last ∧ (bitsinvert(bits) + (¬ boolvar0)').last))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # boolvar ' = 0 +1;% ⊦ # bitsinvert(bits) = # bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + bits0 = bits1 + bits0 ↔ bits = bits1;% ⊦ bitsinvert(boolvar ') = (¬ boolvar)';% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;