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Proof step 2 from proof cleanup-lpt-n-at (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
lpt[m] = mklp(m0, m1, lpf, m2), m < # lpt, m < n, n ≤ # lpt ⊦ cleanup-lpt(lpt, n)[m] = (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m2);lpt[m])

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[m], m0, m1, 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[m], m0, m1, lpf