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Proof step 15 from proof id-idx-freelist-cons (previous step)

# lpt1 = # lpt0, lpt1[n] = mklp(m, m0, lpf, m1), 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 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist ⊦ n < # lpt0 ∧ lpf = LP_FREE ↔ n ∈ freelist, dups(freelist)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# lpt1 = # lpt0, lpt1[n] = mklp(m, m0, lpf, m1), n < # lpt0 → lpt0[n].ref-size = m ∧ lpt0[n].size = m0 ∧ lpt0[n].flags = lpf, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist ⊦ n < # lpt0 ∧ lpf = LP_FREE ↔ n ∈ freelist, dups(freelist)

Rule argument

Additional information

Used simplifier rules: ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s)% ⊦ mklp(n2, n3, lpf, m).size = n3; (s)% ⊦ mklp(n2, n3, lpf, m).ref-size = n2; (s)