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Proof step 11 from proof int2bytes-first-neg (previous step)

~128 ≤ i ∧ i < 128, i2b(i) ' = by ' + bytes, i < 0, 0 ≤ b2i(by), by ∈ i2b(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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ by ∈ by0 ' ↔ by = by0;%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];