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Proof step 4 from proof int2bits-add-high (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ j * 2 ^ i + j0 ⊦ int2bits(j * 2 ^ i + j0) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

Additional information

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