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Proof step 3 from proof bitsand-zeros (previous step)

bits0 = boolvar + bits1, # bits ≤ # bits0 ∨ boolvar ⊦ bits = [], bits0 = [], bitsand(bits, bits0) = bits, true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# bits ≤ 1 + # bits1 ∨ boolvar ⊦ bits = [], bitsand(bits, boolvar ' + bits1) = bits, true ∈ bits

Rule argument

Additional information

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