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Proof step 3 from proof freelist-cons-add (previous step)

lpt[n] = mklp(n0, n1, lpf, n2), n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, dups(freelist), ¬ n ∈ freelist ∧ (n3 < # lpt ∧ lpt[n, mklp(m0, m1, LP_FREE, m3)][n3].flags = LP_FREE ↔ n3 ∈ freelist ∨ n3 = n)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
lpt[n] = mklp(n0, n1, lpf, n2), lpt[n, mklp(m0, m1, LP_FREE, m3)][n3] = mklp(m, m2, lpf0, m4), n < # lpt, ∀ n. n < # lpt ∧ lpt[n].flags = LP_FREE ↔ n ∈ freelist ⊦ lpf = LP_FREE, dups(freelist), ¬ n ∈ freelist ∧ (n3 < # lpt ∧ lpf0 = LP_FREE ↔ n3 ∈ freelist ∨ n3 = n)

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, mklp(m0, m1, LP_FREE, m3)][n3], m, m2, 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, mklp(m0, m1, LP_FREE, m3)][n3], m, m2, lpf0, m4