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Proof step 6 from proof lpt-leb-size-cleanup (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[n] = mklp(m, m0, lpf, m1), lpt[n] = mklp(m2, m3, lpf0, m4), n < # lpt, LEB_SIZE < m0, m3 < m0, m3 ≤ LEB_SIZE, ∀ n. n < # lpt → lpt[n].size ≤ LEB_SIZE ⊦

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] → [lpt[n], m2, m3, lpf0, m4]

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], m2, m3, lpf0, m4