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Proof step 30 from proof int2bytes-len-le-neg-02 (previous step)

0 ≤ 1 + n→i(# int2bytes(i)), 0 ≤ 2 ^ (8 * n→i(# int2bytes(i)) - 1) + i, 8 * (n→i(# int2bytes(i)) - 1) + 1 < 8 * n→i(# int2bytes(i)), 8 * n→i(n) + 1 < 8 * n→i(# int2bytes(i)), # int2bytes(i) ≤ n, 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, i < 0, 1 ≤ # int2bytes(i), 2 ^ (8 * n→i(n)) + i < 0, 1 ≤ n, n ≤ i→n(1 + n→i(# int2bytes(i))) ⊦ n = 0, b2i(int2bytes(i).first) = ~1, int2bytes(i).first = i2b(~1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
n ≤ 1 + # int2bytes(i), 1 ≤ n, 2 ^ (8 * n→i(n)) + i < 0, 1 ≤ # int2bytes(i), i < 0, 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, # int2bytes(i) ≤ n, 8 * n→i(n) + 1 < 8 * n→i(# int2bytes(i)), 8 * (n→i(# int2bytes(i)) - 1) + 1 < 8 * n→i(# int2bytes(i)), 0 ≤ 2 ^ (8 * n→i(# int2bytes(i)) - 1) + i ⊦ int2bytes(i).first = i2b(~1), b2i(int2bytes(i).first) = ~1, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m + n;% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)% ⊦ 0 ≤ i → i→n(i + n→i(n)) = i→n(i) + n;% ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i < 0 ⊦ (* i) ≤ n→i(n);% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;