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Proof step 36 from proof int2bytes-len-le (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: 0 ≤ i ⊦ n→i(n) ≤ (* i) ↔ n ≤ i→n(i) subst: [i, n] → [1 + n→i(# int2bytes(i)), n] rotate: F spec: intnat inst: name: num-n2i-ls-04

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i < 0 ⊦ (* i) ≤ n→i(n);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ 0 ≤ i → i→n(i + n→i(n)) = i→n(i) + n;% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1);% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m ≤ m;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + 0 = i;% ⊦ (* i2) * i + j < (* i1) * i ↔ j < (i1 - i2) * i;% ⊦ i * 0 = 0;% ⊦ i * j * j0 = (i * j) * j0; Used terms: 1 + n→i(# int2bytes(i)), n