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Proof step 3 from proof bits2int-rec-03 (previous step)

bits = boolvar + bits0 ∧ bits ≠ [], boolvar, false ∈ bits ⊦ bits2int(bits + true ') = bits2int(bits) * 2 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
false ∈ bits0 ⊦ bits2uint(bitsinvert(bits0)) + 1 = bits2uint((false ' + bitsinvert(bits0))+1)

Rule argument

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);%bits2uint-one: ⊦ bits2uint(bits + true ') = bits2uint(bits) * 2 + 1;%bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits ≠ [];% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ i * j - i * j0 = i * (j - j0);% ⊦ i * j + i = i * (j + 1);% ⊦ i + (j - i0) = i + j - i0;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ~ i * j = ~(i * j);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + false ')+1 = bits + true;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(true ') = false ';% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ boolvar + bits = boolvar ' + bits;