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Proof step 92 from proof int2bytes-rec-neg-02 (previous step)

# bits2bytes(int2bits(j % 2 ^ (i * 8))) ≥ i→n(i) + 1, 1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), 0 ≤ i, j % 2 ^ (i * 8) < 0, j + 2 ^ (i * 8) < 0, 0 < i ⊦ bits2bytes(int2bits(j % 2 ^ (i * 8)).rest) = lastn(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 93:
1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), i→n(i) < # bits2bytes(int2bits(j % 2 ^ (i * 8))), 0 < i, j + 2 ^ (i * 8) < 0, j % 2 ^ (i * 8) < 0, 0 ≤ i ⊦ int2bits(j % 2 ^ (i * 8)).rest = [], j % 2 ^ (i * 8) = 0, 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;% ⊦ i * j = j * i;