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Proof step 65 from proof int2bytes-rec-pos-any (previous step)

0 < j % 2 ^ (i * 8), # fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, ¬ j % 2 ^ (i * 8) < 2 ^ (n→i(i→n(i * 8) + 1) - 1), 0 < i, 2 ^ (i * 8) ≤ j ⊦ bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, 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 66:
# fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest % 8 = 0, 0 < i, 2 ^ (i * 8) ≤ j, 2 ^ (i * 8) ≤ j % 2 ^ (i * 8), 0 < j % 2 ^ (i * 8) ⊦ bits2bytes(fillsgn(i→n(i * 8) + 1, int2bits(j % 2 ^ (i * 8))).rest) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, 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;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 < j → abs(j) = j;% ⊦ abs(i * j) = abs(i) * abs(j);% ⊦ n→i(i→n(i)) = abs(i);% ⊦ n→i(m + n) = n→i(m) + n→i(n);