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Proof step 2 from proof no-leading-zero-04 (previous step)

true, ∀ bytes, by, i0. i→n(i0) < i→n(i) → true → b2i(by) < 0 → int2bytes(i0) ≠ i2b(~1) ' + by ' + bytes ⊦ b2i(by) < 0 → int2bytes(i) ≠ i2b(~1) ' + by ' + bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
int2bytes(i) = i2b(~1) ' + by ' + bytes, b2i(by) < 0, i2b(~1) ∈ int2bytes(i), ∀ bytes, by, i0. i→n(i0) < i→n(i) → true → b2i(by) < 0 → int2bytes(i0) ≠ i2b(~1) ' + by ' + bytes ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)