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Proof step 164 from proof int2bytes-rec-neg-05 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 165:
i→n(i * 8) + 1 = # int2bits(j % 2 ^ (i * 8)), i→n(i * 8) % 8 = 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8) < i, j % 2 ^ (i * 8) < 0 ⊦ int2bits(j % 2 ^ (i * 8)).rest = [], j % 2 ^ (i * 8) = 0, bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ int2bits(j) ≠ [];% ⊦ n -1 = n - 1;% ⊦ i * j = j * i;