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Proof step 1 from proof int2bits-mod2pot (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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