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Proof step 6 from proof bytes2int-size-06 (previous step)

1 < #(by ' + bytes), b2i(by) < 0, 0 ≤ (bytes2int((by ' + bytes).butlast) * 256 + band(b2i((by ' + bytes).last), 255)) + 2 ^ (8 * n→i(# bytes)), ∀ by, bytes0. # bytes0 < # bytes → 0 ≤ bytes2int(by ' + bytes0) + 2 ^ (8 * n→i(# bytes0)) ∧ b2i(by) < 0 → b2i(by) = ~1 ⊦ b2i(by) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ 2 ^ (8 * n→i(# bytes)) + band(255, b2i(bytes.last)) + 256 * bytes2int(by ' + bytes.butlast), b2i(by) < 0, ∀ by, bytes0. # bytes0 < # bytes → 0 ≤ bytes2int(by ' + bytes0) + 2 ^ (8 * n→i(# bytes0)) ∧ b2i(by) < 0 → b2i(by) = ~1 ⊦ bytes = [], b2i(by) = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;% ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;