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Proof step 1 from proof int2bits-fill-rest (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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