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Proof step 4 from proof int2bits-fill-rest (previous step)

¬(int2bits(j) ≠ [] ∧ # int2bits(j) < i→n(i) + 1), 0 < i, j < 0, 0 < j + 2 ^ (i - 1) ⊦ 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 5:
i→n(i) < # int2bits(j), 0 < j + 2 ^ (i - 1), j < 0, 0 < i ⊦ fillsgn(i→n(i) + 1, int2bits(j)).rest = fillsgn(i→n(i), int2bits(j))

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n + 1 ↔ ¬ n < m;