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

id-idx(lpt0, lpt1) ⊦ lpt-free-unref(lpt1) ↔ lpt-free-unref(lpt0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
id-idx(lpt0, lpt1) ⊦ (∀ n. n < # lpt1 ∧ lpt1[n].flags = LP_FREE → lpt1[n].ref-size = 0) ↔ (∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE → lpt0[n].ref-size = 0)

Rule argument

Additional information

Used simplifier rules: lpt-free-unref: ⊦ lpt-free-unref(lpt) ↔ (∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0);