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Proof step 66 from proof bitsxor-is-orand (previous step)

1 < #(bits0 + boolvar0 ') ∧ 1 < #(bits2 + boolvar '), ∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bits0 = [], bits2 = [], bits2int(bitsxor((bits0 + boolvar0 ').butlast, (bits2 + boolvar ').butlast) + (¬((bits0 + boolvar0 ').last ↔ (bits2 + boolvar ').last))) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)'), bitsand(bitsinvert(bits0), bits2) = [], bitsand(bitsinvert(bits2), bits0) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 67:
∀ bits, bits1. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ bits2int(bitsxor(bits, bits1)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits1), bitsand(bitsinvert(bits1), bits))) ⊦ bitsand(bitsinvert(bits0), bits2) = [], bitsand(bitsinvert(bits2), bits0) = [], bits0 = [], bits2 = [], bits2int(bitsxor(bits0, bits2) + (¬(boolvar0 ↔ boolvar))') = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2), bitsand(bitsinvert(bits2), bits0)) + (¬ boolvar0 ∧ boolvar ∨ ¬ boolvar ∧ boolvar0)')

Rule argument

Additional information

Used simplifier rules: ⊦ # 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;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;