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Proof step 2 from proof sis-03 (previous step)

i + j ≤ 32767, 0 ≤ i, 0 ≤ j ⊦ s2i(i2s(i)) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i ∈short → s2i(i2s(i)) = i
Proof step 4:
¬ i ∈short, i + j ≤ 32767, 0 ≤ i, 0 ≤ j ⊦ s2i(i2s(i)) = i

Rule argument

name: sis-in rotate: F subst: [i] → [i]

Additional information

Used terms: i