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Proof step 13 from proof uint2bits-add-high-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i→n(i) + 1 < # int2bits(j0), 0 ≤ i, 0 ≤ j0 + 2 ^ i, 2 ^ (i - 1) + j0 < 0, 0 < i, j0 < 0, 0 < j, 1 < j ⊦ fillsgn(i→n(i) + 1, sgn(int2bits(j0).rest)).rest = fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;