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Proof step 10 from proof band-011x-eq (previous step)

j < 0, 0 ≤ i ⊦ (i - j) % (i + 1) + (j + 1) % 2 ^ n→i(# int2bits(i) - 1) + 2 ^ n→i(# int2bits(i) - 1) = i + 1, ∀ i0. i ≠ 2 ^ i0 - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 ≤ i0, j < 0 ⊦ (2 ^ i0 - 1 + j) % 2 ^ i0 + (j + 1) % 2 ^ i0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i = i + j ↔ j = 0;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;% ⊦ i - j - i0 = i - j + i0;% ⊦ n→i(i→n(i)) = abs(i);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # mklist(boolvar, n) = n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;%0 ≤ i ⊦ int2bits(2 ^ i - 1) = false ' + mklist(true, i→n(i));% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < 0 → 2 ^ i = 0;