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Proof step 5 from proof int2bytes-first-neg (previous step)

int2bytes(i) = by ' + bytes ∧ by ∈ int2bytes(i) ∧ 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 ⊦ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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 ⊦

Proof step 5 from proof int2bytes-first-neg (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information