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Proof step 3 from proof int2bytes-first-neg (previous step)

int2bytes(i) = by + bytes ∧ int2bytes(i) ≠ [], 0 ≤ b2i(by), i < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
int2bytes(i) = by ' + bytes, by ∈ int2bytes(i), 0 ≤ b2i(by), i < 0 ⊦

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes = bytes0 → by ∈ bytes0; (forward)% ⊦ by + bytes = by ' + bytes;