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Proof step 50 from proof bitsxor-is-orand (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
∀ 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), bits.butlast) + (¬ boolvar ∧ bits.last)'))

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;% ⊦ # bitsinvert(bits) = # bits;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;