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Proof step 29 from proof int2bytes-len-le-neg-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ n ≤ m → n→i(m - n) = n→i(m) - n→i(n)
Proof step 31:
1 ≤ # int2bytes(i), int2bytes(i).first = i2b(~1), 2 ^ (8 * (n→i(# int2bytes(i)) - n→i(1)) - 1) + i < 0, 0 ≤ 2 ^ (8 * (n→i(# int2bytes(i)) - n→i(1))) + i, 2 ≤ # int2bytes(i), 0 ≤ b2i(int2bytes(i).rest.first) ⊦ int2bytes(i).rest = [], # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n) - 1) + i < 0

Rule argument

seq: ⊦ n ≤ m → n→i(m - n) = n→i(m) - n→i(n) subst: [n, m] → [1, # int2bytes(i)] rotate: F spec: intnat inst: name: n2i-sub

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ 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; Used terms: 1, # int2bytes(i)