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Proof step 10 from proof int2bytes-first-neg-01 (previous step)

i2b(i) = 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 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 12:
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 ⊦

Rule argument

seq: ⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j)) subst: [j, i] → [0, i] rotate: F spec: byte inst: name: i2b-equal

Additional information

Used terms: 0, i