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Proof step 26 from proof bits2int-lasteq (previous step)

boolvar ∧ (boolvar0 ' + bits1).first ∧ boolvar0 ' + bits1 ≠ [] ⊦ bits0 = [], ~(bits2uint(bitsinvert((boolvar0 ' + bits1).rest)) * 2 + 1) = bits2int(bits0 + boolvar ') ↔ bits2int(boolvar0 ' + bits1) = bits2int(bits0), boolvar ∧ ¬ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
⊦ bits0 = [], ~1 = 2 * bits2uint(bitsinvert(bits1)) + bits2int(bits0 + true ') ↔ bits2uint((false ' + bitsinvert(bits1))+1) + bits2int(bits0) = 0

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ boolvar ' + bits ≠ [];% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;