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Proof step 5 from proof id-idx-freelist-cons (previous step)

# lpt1 = # lpt0, ∀ n. n < # lpt1 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ (∀ n. n < # lpt1 ∧ lpt1[n].flags = LP_FREE ↔ n ∈ freelist) ↔ (∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist), dups(freelist)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# 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].flags = LP_FREE ↔ n ∈ freelist) ↔ (∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist), dups(freelist)

Proof step 5 from proof id-idx-freelist-cons (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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