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Proof step 3 from proof id-idx-cleanup-lpt-backwards-fwd (previous step)

lpt0[n] = mklp(m0, m1, lpf0, m2), lpt1[n] = mklp(n0, n1, lpf1, n2), n < # lpt0, id-idx(cleanup-lpt(lpt0, n), lpt1) ⊦ m0 = n0 ∧ m1 = n1 ∧ lpf0 = lpf1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ id-idx(lpt1, lpt0) ↔ # lpt0 = # lpt1 ∧ (∀ n. n < # lpt0 → lpt1[n].ref-size = lpt0[n].ref-size ∧ lpt1[n].size = lpt0[n].size ∧ lpt1[n].flags = lpt0[n].flags)
Proof step 5:
# lpt1 = # cleanup-lpt(lpt0, n), lpt1[n] = mklp(n0, n1, lpf1, n2), lpt0[n] = mklp(m0, m1, lpf0, m2), n < # lpt0, ∀ n0. n0 < # lpt1 → cleanup-lpt(lpt0, n)[n0].ref-size = lpt1[n0].ref-size ∧ cleanup-lpt(lpt0, n)[n0].size = lpt1[n0].size ∧ cleanup-lpt(lpt0, n)[n0].flags = lpt1[n0].flags ⊦ m0 = n0 ∧ m1 = n1 ∧ lpf0 = lpf1

Rule argument

spec: index-array, inst: gc-heap-base name: id-idx seq: ⊦ id-idx(lpt1, lpt0) ↔ # lpt0 = # lpt1 ∧ (∀ n. n < # lpt0 → lpt1[n].ref-size = lpt0[n].ref-size ∧ lpt1[n].size = lpt0[n].size ∧ lpt1[n].flags = lpt0[n].flags) subst: [lpt0, lpt1] → [lpt1, cleanup-lpt(lpt0, n)] flags: <all paths>

Additional information

Used terms: lpt1, cleanup-lpt(lpt0, n)