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Proof step 7 from proof band-011x-is-unsigned (previous step)

j < 0 ∧ ¬ false ∈ int2bits(2 ^ i - 1).rest ∧ 0 ≤ 2 ^ i - 1, 0 ≤ j + 2 ^ (i - 1), j < 0, 0 < i ⊦ (j + 1) % ((2 ^ i - 1) + 1) + (2 ^ i - 1) = bits2uint(fillsgn(i→n(i), int2bits(j)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ i, 0 < i, j < 0, 0 ≤ j + 2 ^ (i - 1) ⊦ (j + 1) % 2 ^ i + 2 ^ i = bits2uint(fillsgn(i→n(i), int2bits(j))) + 1

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)% ⊦ ¬ false ∈ mklist(true, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;%0 ≤ i ⊦ int2bits(2 ^ i - 1) = false ' + mklist(true, i→n(i));% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i) = j;