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Proof step 2 from proof id-idx-freelist-cons (previous step)

id-idx(lpt0, lpt1) ⊦ freelist-cons(lpt1, freelist) ↔ freelist-cons(lpt0, freelist)

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 ↔ n ∈ freelist) ↔ (∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist), dups(freelist)

Rule argument

Additional information

Used simplifier rules: freelist-cons: ⊦ freelist-cons(lpt, freelist) ↔ ¬ dups(freelist) ∧ (∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist);