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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
lpt[n] = mklp(m, m0, lpf, m1), n < # lpt ⊦ (m = 0 ∧ lpf ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, m1);lpt[n]) = lpt.map(λ lp. (lp.ref-size = 0 ∧ lp.flags ≠ LP_FREE ⊃ mklp(0, 0, LP_FREE, lp.gcheapidx);lp))[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] → [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