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Proof step 12 from proof lpt-free-unref-replay-lpt (previous step)

replay-lpt(lpt, nx, n)[n0].flags = LP_FREE, n0 < # lpt, n ≤ # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ replay-lpt(lpt, nx, n)[n0].ref-size = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
replay-lpt(lpt, nx, n)[n0] = mklp(m, m0, lpf, m1), lpf = LP_FREE, n0 < # lpt, n ≤ # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[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] → [replay-lpt(lpt, nx, n)[n0], 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: replay-lpt(lpt, nx, n)[n0], m, m0, lpf, m1