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Proof step 25 from proof no-leading-zero-01 (previous step)

int2bytes(i1 / 256) = i2b(0) ' + (by ' + bytes).butlast, int2bytes((i1 + 255) % 256).last = (by ' + bytes).last, 0 ≤ b2i(by), i1 < ~383, i1 < 0, i2b(0) ∈ int2bytes(i1 / 256), ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = i2b(0) ' + by ' + bytes ∧ 0 ≤ b2i(by) ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ bytes = [], (by ' + bytes).last ≠ i2b(0) ∧ (by ' + bytes).last ≠ by ∧ ¬ (by ' + bytes).last ∈ bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
int2bytes(i1 / 256) = i2b(0) ' + by ' + bytes.butlast, int2bytes((i1 + 255) % 256).last = bytes.last, 0 ≤ b2i(by), i1 < ~383, i1 < 0, i2b(0) ∈ int2bytes(i1 / 256), ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = i2b(0) ' + by ' + bytes ∧ 0 ≤ b2i(by) ∧ i2b(0) ∈ int2bytes(i0) → false ⊦ bytes = []

Rule argument

Additional information

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