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Proof step 7 from proof and-first (previous step)

bitsand(boolvar ' + bits1, boolvar0 ' + bits) = boolvar1 + bits0 ∧ bitsand(boolvar ' + bits1, boolvar0 ' + bits) ≠ [] ⊦ boolvar1 ↔ boolvar ∧ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
bitsand(boolvar ' + bits1, boolvar0 ' + bits) = boolvar1 ' + bits0 ⊦ boolvar1 ↔ boolvar ∧ boolvar0

Rule argument

Additional information

Used simplifier rules: and-notempty-06: ⊦ bitsand(boolvar ' + bits, boolvar0 ' + bits0) ≠ [];% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;