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Proof step 8 from proof bytes2int-size-05 (previous step)

band(255, b2i((by ' + bytes).last)) + 256 * bytes2int(i2b(~1) ' + (by ' + bytes).butlast) + 2 ^ ((n→i(# bytes) + 1) * 8) < 0, 0 ≤ b2i(by), ∀ by, bytes0. # bytes0 < # bytes → 2 ^ ((n→i(# bytes0) + 1) * 8) + bytes2int(i2b(~1) ' + by ' + bytes0) < 0 ∧ 0 ≤ b2i(by) → false ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
2 ^ ((n→i(# bytes) + 1) * 8) + 256 * bytes2int(i2b(~1) ' + by ' + bytes.butlast) + band(255, b2i(bytes.last)) < 0, 0 ≤ b2i(by), ∀ by, bytes0. # bytes0 < # bytes → 2 ^ ((n→i(# bytes0) + 1) * 8) + bytes2int(i2b(~1) ' + by ' + bytes0) < 0 ∧ 0 ≤ b2i(by) → false ⊦ bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;