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Proof step 6 from proof and-true (previous step)

bits = boolvar + bits1, bitsand(bits, bits0) = true ' ⊦ bits0 = [], bits = [], bits = bitsand(bits, bits0) ∧ bits0 = bitsand(bits, bits0), bits ≠ boolvar ' ∧ bits0 ≠ bits0.first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
bitsand(boolvar ' + bits1, bits0) = true ' ⊦ bits0 = [], boolvar ' + bits1 = true ' ∧ bits0 = true ', bits0 ≠ bits0.first ' ∧ bits1 ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ bits + [] = bits;% ⊦ boolvar ' = boolvar0 ' + bits ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;