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Proof step 20 from proof id-idx-lpt-free-unref (previous step)

# lpt1 = # lpt0, lpt0[n].flags = LP_FREE, lpt1[n] = mklp(lpt0[n].ref-size, lpt0[n].size, LP_FREE, m1), n < # lpt0, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE → lpt0[n].ref-size = 0 ⊦ lpt0[n].ref-size = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
# lpt1 = # lpt0, lpt1[n] = mklp(m, m0, LP_FREE, m1), lpt0[n] = mklp(m, m0, lpf, m2), lpf = LP_FREE, n < # lpt0, ∀ n. n < # lpt0 ∧ lpt0[n].flags = LP_FREE → lpt0[n].ref-size = 0 ⊦ 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] → [lpt0[n], m, m0, lpf, m2]

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], m, m0, lpf, m2