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Proof step 12 from proof int2bytes-first-neg-01 (previous step)

b2i(i2b(i)) = b2i(i2b(0)), ~128 ≤ i, i < 0, ∀ 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

Proof step 13:
b2i(i2b(i)) = 0, ~128 ≤ i, i < 0, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ b2i(i2b(0)) = 0;