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Proof step 5 from proof cleanup-lpt-n-read-above (previous step)

n0 < # lpt, n0 < m, m < # lpt, n0 + 1 < # lpt, ∀ lpt, m. n0 < # lpt ∧ m < # lpt ∧ n0 ≤ m → cleanup-lpt(lpt, n0)[m] = lpt[m] ⊦ (lpt[n0].ref-size = 0 ∧ lpt[n0].flags ≠ LP_FREE ⊃ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, lpt[n0].gcheapidx)];cleanup-lpt(lpt, n0))[m] = lpt[m]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
lpt[n0] = mklp(m0, m1, lpf, m2), n0 + 1 < # lpt, m < # lpt, n0 < m, n0 < # lpt, ∀ lpt, m. n0 < # lpt ∧ m < # lpt ∧ n0 ≤ m → cleanup-lpt(lpt, n0)[m] = lpt[m] ⊦ (m0 = 0 ∧ lpf ≠ LP_FREE ⊃ cleanup-lpt(lpt, n0)[n0, mklp(0, 0, LP_FREE, m2)];cleanup-lpt(lpt, n0))[m] = 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[n0], 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[n0], m0, m1, lpf