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Proof step 8 from proof int2bytes-len-eq-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
int2bytes(i) = i2b(0) ' + bytes, 0 ≤ i, 0 ≤ b2i(bytes.first), i2b(0) ∈ int2bytes(i) ⊦ # 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) ≠ [];% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ by + bytes = by ' + bytes;