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Proof step 71 from proof int2bytes-rec-pos-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 72:
i→n(i * 8) + 1 = # int2bits(j % 2 ^ (i * 8)), # int2bits(j % 2 ^ (i * 8)).rest % 8 = 0, 0 < i, 2 ^ (i * 8) ≤ j ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = fix(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest, int2bits(j % 2 ^ (i * 8)).rest = []

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;% ⊦ mklist(boolvar, 0) = [];% ⊦ m - m = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n - n + n1 = m - n1;% ⊦ i * j = j * i;% ⊦ m ≤ n → (n - m = 0 ↔ n = m);