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Proof step 55 from proof bitsxor-is-orand (previous step)

¬(1 < #(bitsand(bitsinvert(bits0), bits2) + (¬ boolvar0 ∧ boolvar)') ∧ 1 < #(bitsand(bitsinvert(bits2), bits0) + (¬ boolvar ∧ boolvar0)')), ∀ 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 ', bits2 + boolvar ')) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2) + (¬ boolvar0 ∧ boolvar)', bitsand(bitsinvert(bits2), bits0) + (¬ boolvar ∧ boolvar0)'))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
∀ 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 ', bits2 + boolvar ')) = bits2int(bitsor(bitsand(bitsinvert(bits0), bits2) + (¬ boolvar0 ∧ boolvar)', bitsand(bitsinvert(bits2), bits0) + (¬ boolvar ∧ boolvar0)')), bitsand(bitsinvert(bits0), bits2) ≠ [] ∧ bitsand(bitsinvert(bits2), bits0) ≠ []

Rule argument

Additional information

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