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Proof step 62 from proof int2bytes-len-eq-pos (previous step)

1 ≤ # int2bytes(i) - 1, int2bytes(i).first = i2b(0), 2 ^ (8 * (1 + (n→i(# int2bytes(i) - 1) - n→i(1))) - 1) ≤ i, n→i(n) ≤ 1 + n→i(# int2bytes(i) - 1), 2 ^ ((n→i(n) - 1) * 8 - 1) ≤ i, i < 2 ^ ((n→i(n) - 1) * 8), i < 2 ^ (8 * n→i(# int2bytes(i) - 1)), 2 ≤ # int2bytes(i), 0 ≤ i, 1 + (# int2bytes(i) - 1 - 1) < n ⊦ int2bytes(i).rest = [], # int2bytes(i) = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 63:
int2bytes(i).first = i2b(0), 2 ^ (8 * n→i(# int2bytes(i) - 1) - 1) < 2 ^ ((n→i(n) - 1) * 8), 2 ^ (8 * n→i(# int2bytes(i) - 1) - 1) ≤ i, 1 + (# int2bytes(i) - 1 - 1) < n, 0 ≤ i, 2 ≤ # int2bytes(i), i < 2 ^ (8 * n→i(# int2bytes(i) - 1)), i < 2 ^ ((n→i(n) - 1) * 8), 2 ^ ((n→i(n) - 1) * 8 - 1) ≤ i, n→i(n) ≤ 1 + n→i(# int2bytes(i) - 1) ⊦ # int2bytes(i) = n, int2bytes(i).rest = []

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ n→i(n);% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 2 ^ (i - 1) < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ i * j = j * i;