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Proof step 3 from proof id-idx-write-both (previous step)

id-idx(lpt0, lpt1) ⊦ id-idx(lpt0[n, mklp(n0, n1, lpf, n2)], lpt1[n, mklp(n0, n1, lpf, m2)])

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 = # lpt0, ∀ n. n < # lpt1 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ id-idx(lpt0[n, mklp(n0, n1, lpf, n2)], lpt1[n, mklp(n0, n1, lpf, m2)])

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, lpt0] flags: <all paths>

Additional information

Used terms: lpt1, lpt0