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Proof step 6 from proof bits2int-lasteq (previous step)

bits = boolvar0 + bits1 ⊦ bits = [], bits0 = [], bits2int(bits + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2int(bits) = bits2int(bits0), boolvar ∧ ¬ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ bits0 = [], bits2int((boolvar0 ' + bits1) + boolvar ') = bits2int(bits0 + boolvar ') ↔ bits2int(boolvar0 ' + bits1) = bits2int(bits0), boolvar ∧ ¬ boolvar0

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);% ⊦ ~ i = j ↔ i + j = 0;% ⊦ bitsinvert(boolvar ') = (¬ boolvar)';% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;