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Proof step 4 from proof byte-short-neq (previous step)

i2s(i) = i2s(j) ⊦ i2b(cast(8, i)) = i2b(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i2s(i) = i2s(j) ↔ s2i(i2s(i)) = s2i(i2s(j))
Proof step 6:
s2i(i2s(i)) = s2i(i2s(j)) ⊦ i2b(cast(8, i)) = i2b(j)

Rule argument

name: short-inj rotate: F subst: [j, i] → [j, i]

Additional information

Used terms: j, i