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Proof step 14 from proof int2bytes-zero (previous step)

i2b(i) = i2b(0), ~128 ≤ i, i ≤ 127 ⊦ i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

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 simplifier rules: ⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i;% ⊦ b2i(i2b(0)) = 0; Used terms: 0, i