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Proof step 52 from proof bitsxor-is-orand (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 53:
∀ 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)'))

Rule argument

Additional information

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