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Proof step 8 from proof int2bytes-first-neg (previous step)

¬ i < ~ 128, int2bytes(i) = by ' + bytes, i < 0, 0 ≤ b2i(by), by ∈ int2bytes(i), ∀ 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 9:
int2bytes(i) = by ' + bytes, i < 0, 0 ≤ b2i(by), by ∈ int2bytes(i), ~128 ≤ i, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;