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Proof step 2 from proof int2bits-rec-2pot-pos (previous step)

j = i0 * 2 ^ i + j0 ∧ 0 ≤ j0 ∧ j0 < 2 ^ i ∧ 0 ≤ i0 ∧ 0 < 2 ^ i ∧ 0 ≤ j ⊦ 2 ^ i ≤ j ∧ 0 < i → int2bits(j) = int2bits(i0) + fillsgn(i→n(i + 1), int2bits(j0)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≤ j0, 0 < i, j0 < 2 ^ i, 2 ^ i ≤ i0 * 2 ^ i + j0, 0 ≤ i, 0 ≤ i0, 0 < i0 ⊦ int2bits(i0 * 2 ^ i + j0) = int2bits(i0) + fillsgn(i→n(i) + 1, int2bits(j0)).rest

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i < i + j ↔ 0 < j;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1;