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Proof step 3 from proof bits2int-split-01 (previous step)

bits = boolvar + bits1, boolvar ⊦ ~1 = 2 ^ n→i(# bits0) * bits2uint(bitsinvert(bits)) + bits2uint(bitsinvert(bits0)) + bits2int(bits + bits0), bits = [], bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ bits0 = [], bits2uint((false ' + bitsinvert(bits1) + bitsinvert(bits0))+1) = (bits2uint(bitsinvert(bits1)) * 2 ^ n→i(# bits0) + bits2uint(bitsinvert(bits0))) + 1

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);%bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ boolvar ' + bits ≠ [];% ⊦ i + ~1 = i - 1;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + ~ j = i - j;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ boolvar + bits = boolvar ' + bits;