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Proof step 2 from proof lpt-leb-size-id-idx (previous step)

id-idx(lpt0, lpt1) ⊦ lpt-leb-size(lpt1) ↔ lpt-leb-size(lpt0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
id-idx(lpt0, lpt1) ⊦ (∀ n. n < # lpt1 → lpt1[n].size ≤ LEB_SIZE) ↔ (∀ n. n < # lpt0 → lpt0[n].size ≤ LEB_SIZE)

Rule argument

Additional information

Used simplifier rules: lpt-leb-size: ⊦ lpt-leb-size(lpt) ↔ (∀ n. n < # lpt → lpt[n].size ≤ LEB_SIZE);