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Proof step 84 from proof int2bytes-rec-pos-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 85:
# int2bits(j % 2 ^ (i * 8)).rest % 8 = 0, # int2bits(j % 2 ^ (i * 8)) = i→n(i * 8) + 1, i→n(i) < # bits2bytes(int2bits(j % 2 ^ (i * 8))), 2 ^ (i * 8) ≤ j, 0 < i ⊦ int2bits(j % 2 ^ (i * 8)).rest = [], bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(i→n(i) + 1, bits2bytes(int2bits(j % 2 ^ (i * 8)))).rest

Rule argument

Additional information

Used simplifier rules: bits2bytes-empty: ⊦ bits2bytes(bits) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ n ≥ m ↔ m ≤ n;