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Proof step 16 from proof id-idx-freelist-cons (previous step)

# 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)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
# lpt1 = # lpt0, lpt1[n] = mklp(m, m0, lpf, m1), lpt0[n] = mklp(m2, m3, lpf0, m4), n < # lpt0 → m2 = m ∧ m3 = m0 ∧ lpf0 = lpf, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE ↔ n ∈ freelist ⊦ n < # lpt0 ∧ lpf = LP_FREE ↔ n ∈ freelist, dups(freelist)

Rule argument

spec: lprops name: elim seq: ⊦ ((m = lp.ref-size ∧ m0 = lp.size) ∧ lpf = lp.flags) ∧ m1 = lp.gcheapidx ↔ lp = mklp(m, m0, lpf, m1) subst: [lp, m, m0, lpf, m1] → [lpt0[n], m2, m3, lpf0, m4]

Additional information

Used simplifier rules: ⊦ ((m = lp.ref-size ∧ m0 = lp.size) ∧ lpf = lp.flags) ∧ m1 = lp.gcheapidx ↔ lp = mklp(m, m0, lpf, m1); (elim) Used terms: lpt0[n], m2, m3, lpf0, m4