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Proof step 1 from proof int2bits-rec-2pot-neg-02 (conclusion)

⊦ j ≤ ~(2 ^ i) ∧ 0 < i ∧ j % 2 ^ i ≠ 0 → int2bits(j) = int2bits(j / 2 ^ i - 1) + fillsgn(i→n(i + 1), int2bits(j % 2 ^ i)).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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