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Proof step 7 from proof int2bits-mod2pot (previous step)

0 ≤ i ∧ 2 ^ i ≤ j * 2 ^ i + j0, 0 < j, 0 ≤ j0, j0 < 2 ^ i, 0 ≤ i, 0 < 2 ^ i, 0 ≤ j ⊦ uint2bits((j * 2 ^ i + j0) / 2 ^ i) + fillsgn(i→n(i + 1), uint2bits((j * 2 ^ i + j0) % 2 ^ i)).rest = int2bits(j) + fillsgn(i→n(i) + 1, int2bits(j0)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ j, 0 ≤ i, j0 < 2 ^ i, 0 ≤ j0, 0 < j, 2 ^ i ≤ j * 2 ^ i + j0, 0 < j * 2 ^ i + j0, 0 < 2 ^ i ⊦ uint2bits(j) + fillsgn(i→n(i) + 1, uint2bits((j * 2 ^ i + j0) % 2 ^ i)).rest = int2bits(j) + 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 ∧ j ≤ j0 → i < j0; (forward)% ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ i0 ∧ i0 < j ∧ 0 ≤ i → (i * j + i0) / j = i;