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Proof step 3 from proof and-size-one-01 (previous step)

bits = boolvar + bits2 ∧ bits ≠ [], # bits = # bits1 + 1, boolvar ⊦ bits1 = [], bits0 = [], bitsand(bits, bits0 + bits1) = bits0 + bitsand(bits2, bits1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# bits2 = # bits1 ⊦ bits1 = [], bits0 = [], bitsand(true ' + bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1)

Rule argument

Additional information

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