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Proof step 59 from proof int2bytes-len-eq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
int2bytes(i).first = i2b(0), 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1 - 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

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;%n ≤ m ⊦ (m - n) + m0 ≤ n0 ↔ m + m0 ≤ n0 + n;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;% ⊦ i ≥ 0 → ((* i) + n→i(n) ≤ n→i(m) ↔ i→n(i) + n ≤ m);% ⊦ i ≥ 0 → ((* i) + n→i(n) < n→i(m) ↔ i→n(i) + n < m);% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ i + 1 ≤ j ↔ i < j;% ⊦ 0 < i → (i * j ≤ i * i0 ↔ j ≤ i0);% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ i - j < i0 ↔ i < j + i0;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ 0 ≤ n→i(n);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ n -1 = n - 1;