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Proof step 7 from proof bytes2int-size-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;