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Proof step 23 from proof int2bytes-len-eq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
int2bytes(i).first = i2b(0), # int2bytes(i) < 2, 0 ≤ i, i < 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1 - 1))) ⊦ int2bytes(i).rest = [], # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ m < n → n ≠ 0; (forward)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bytes = 0 ↔ bytes = [];