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Proof step 29 from proof int2bytes-len-le (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
(n→i(n) - 1) * 8 + 1 < 8 * (1 + n→i(# int2bytes(i) - 1)), 8 * n→i(# int2bytes(i) - 1) + 1 < 8 * (1 + n→i(# int2bytes(i) - 1)), 2 ^ ((n→i(n) - 1) * 8) ≤ i, # int2bytes(i) < n, 2 ^ (8 * n→i(# int2bytes(i) - 1)) ≤ i, 0 ≤ i, i < 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1)) - 1) ⊦ int2bytes(i).first = i2b(0)

Rule argument

Additional information

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