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Proof step 17 from proof lpt-free-unref-cleanup-lpt (previous step)

(lpt[n].ref-size = 0 ∧ lpt[n].flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lpt[n].gcheapidx);lpt[n]) = mklp(m, m0, LP_FREE, m1), n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE → lpt[n].ref-size = 0 ⊦ m = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
(m2 = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m4);lpt[n]) = mklp(m, m0, LP_FREE, m1), lpt[n] = mklp(m2, m3, lpf, m4), 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] → [lpt[n], m1, m2, lpf]

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: lpt[n], m1, m2, lpf