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Proof step 10 from proof int2bytes-rec-pos-any (previous step)

0 < j ∧ 0 < i * 8 → ¬ i→n(i * 8) + 1 < # int2bits(j % 2 ^ (i * 8)), 2 ^ (i * 8) ≤ j, 0 < i ⊦ bits2bytes(int2bits(j / 2 ^ (i * 8)) + fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = int2bytes(j / 2 ^ (i * 8)) + fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest, # fillsgn(i→n(i * 8 + 1), int2bits(j % 2 ^ (i * 8))).rest % 8 = 0 ∧ fillsgn(i→n(i * 8 + 1), int2bits(j % 2 ^ (i * 8))).rest ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
2 ^ (i * 8) ≤ j, 0 < i, # int2bits(j % 2 ^ (i * 8)) ≤ i→n(i * 8) + 1, 0 < j ⊦ bits2bytes(int2bits(j / 2 ^ (i * 8)) + fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = int2bytes(j / 2 ^ (i * 8)) + fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest, # fillsgn(i→n(i * 8 + 1), int2bits(j % 2 ^ (i * 8))).rest % 8 = 0 ∧ fillsgn(i→n(i * 8 + 1), int2bits(j % 2 ^ (i * 8))).rest ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ m + n = n + m;% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ 2 ^ i ≤ 0 ↔ i < 0;% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ ¬ i < j ↔ j ≤ i;