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Proof step 16 from proof id-idx-lpt-free-unref (previous step)

# lpt1 = # lpt0, lpt1[n].flags = LP_FREE, n < # 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 ∧ lpt0[n].flags = LP_FREE → lpt0[n].ref-size = 0 ⊦ lpt1[n].ref-size = 0

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), lpf = LP_FREE, n < # lpt0, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE → lpt0[n].ref-size = 0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ m = 0

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] → [lpt1[n], m, m0, lpf, m1]

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: lpt1[n], m, m0, lpf, m1