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Proof step 99 from proof int2bytes-rec-pos-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 100:
# int2bits(j % 2 ^ (i * 8)) = i→n(i * 8) + 1, # int2bits(j % 2 ^ (i * 8)).rest % 8 = 0, i→n(i) + 1 = # bits2bytes(int2bits(j % 2 ^ (i * 8))), (i→n(i * 8) + 1) % 8 = 0, 0 < i, 2 ^ (i * 8) ≤ j ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = bits2bytes(int2bits(j % 2 ^ (i * 8))).rest, int2bits(j % 2 ^ (i * 8)).rest = []

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 ≤ (* j) ↔ i0 ≤ j / i);%0 ≤ i ⊦ i→n(i) ≤ (* m) ↔ i ≤ n→i(m);% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ i * j * j0 = (i * j) * j0;%n < n0 ⊦ m + (* n) ≤ (* n0) ↔ m ≤ n0 - n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i * j = j * i;% ⊦ ¬ m < n ↔ n ≤ m;