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Proof step 2 from proof lpt-free-unref-cleanup-lpt (previous step)

⊦ (∀ n. n < # lpt ∧ lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n].flags = LP_FREE → lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n].ref-size = 0) ↔ (∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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