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Proof step 9 from proof int2bytes-first-neg-01 (previous step)

~128 ≤ i ∧ i < 128, i2b(i) ' = i2b(0) ' + bytes, i2b(0) ∈ i2b(i) ', i < 0, ~128 ≤ i, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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