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Proof step 14 from proof int2bytes-first-neg (previous step)

i < ~ 128, int2bytes((i - 255) / 256) + int2bytes(i % 256).last ' = by ' + bytes, i < 0, 0 ≤ b2i(by), by ∈ int2bytes((i - 255) / 256) + int2bytes(i % 256).last ', ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
int2bytes((i - 255) / 256) + int2bytes(i % 256).last ' = by ' + bytes, i < ~128, 0 ≤ b2i(by), i < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ int2bytes(i % 256).last ≠ by ∧ ¬ int2bytes(i % 256).last ∈ bytes, ¬ by ∈ int2bytes((i - 255) / 256) ∧ by ≠ int2bytes(i % 256).last

Rule argument

Additional information

Used simplifier rules: ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ bytes + by ' = bytes0 → by ∈ bytes0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ∈ by0 ' ↔ by = by0;% ⊦ by ∈ bytes + bytes0 ↔ ¬ ¬(by ∈ bytes ∨ by ∈ bytes0);