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Proof step 4 from proof lpt-free-unref-cleanup-lpt (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
lpt[n] = mklp(m, m0, lpf, m1), lpf = LP_FREE, n < # lpt, ∀ n. n < # lpt ∧ lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n].flags = LP_FREE → lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[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], m, m0, 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], m, m0, lpf