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Proof step 37 from proof int2bytes-len-le-neg-01 (previous step)

int2bytes(i) = by + bytes ∧ int2bytes(i) ≠ [], by = i2b(~1), 0 ≤ 2 ^ ((n→i(# bytes.rest) + 1) * 8) + bytes2int(i2b(~1) ' + bytes.first ' + bytes.rest), bytes2int(i2b(~1) ' + bytes.first ' + bytes.rest) + 2 ^ (8 * (n→i(# bytes.rest) + 1) - 1) < 0, 0 ≤ b2i(bytes.first) ⊦ i2b(~1) ' + bytes.first ' + bytes.rest = int2bytes(i), n ≤ # int2bytes(i) ↔ ¬ 0 ≤ i + 2 ^ (8 * (n→i(n) - 1) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
int2bytes(i) = i2b(~1) ' + bytes, i2b(~1) ∈ int2bytes(i), 0 ≤ b2i(bytes.first), bytes2int(i2b(~1) ' + bytes.first ' + bytes.rest) + 2 ^ (8 * (n→i(# bytes.rest) + 1) - 1) < 0, 0 ≤ 2 ^ ((n→i(# bytes.rest) + 1) * 8) + bytes2int(i2b(~1) ' + bytes.first ' + bytes.rest) ⊦ n ≤ # int2bytes(i) ↔ ¬ 0 ≤ i + 2 ^ (8 * (n→i(n) - 1) - 1), i2b(~1) ' + bytes.first ' + bytes.rest = int2bytes(i)

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ i * j * j0 = (i * j) * j0;%i < 0 ⊦ (* i) ≤ n→i(n);% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);% ⊦ 2 ^ (i - 1) < 2 ^ i ↔ 0 ≤ i;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ by + bytes = by ' + bytes;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i * j = j * i;% ⊦ i + j = j + i;