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Proof step 5 from proof bufs-log-only-id-idx (previous step)

id-idx(lpt0, lpt1) ⊦ m < # lpt1 ∧ (nx ≠ [] ∧ m = nx.last ∧ lpt1[nx.last].flags = LP_GROUP_NODES ∨ lpt1[m].flags = LP_INDEX_NODES) ↔ m < # lpt0 ∧ (nx ≠ [] ∧ m = nx.last ∧ lpt0[nx.last].flags = LP_GROUP_NODES ∨ lpt0[m].flags = LP_INDEX_NODES)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
lpt1[nx.last] = mklp(m0, m1, lpf, m2), id-idx(lpt0, lpt1) ⊦ m < # lpt1 ∧ (nx ≠ [] ∧ m = nx.last ∧ lpf = LP_GROUP_NODES ∨ lpt1[m].flags = LP_INDEX_NODES) ↔ m < # lpt0 ∧ (nx ≠ [] ∧ m = nx.last ∧ lpt0[nx.last].flags = LP_GROUP_NODES ∨ lpt0[m].flags = LP_INDEX_NODES)

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] → [lpt1[nx.last], m0, m1, lpf, m2]

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: lpt1[nx.last], m0, m1, lpf, m2