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Proof step 6 from proof lpt-leb-size-id-idx (previous step)

# lpt1 = # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ (∀ n. n < # lpt0 → lpt1[n].size ≤ LEB_SIZE) ↔ (∀ n. n < # lpt0 → lpt0[n].size ≤ LEB_SIZE)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information