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Proof step 3 from proof bufs-log-only-id-idx (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
bleb = buffered(m), id-idx(lpt0, lpt1) ⊦ m < # lpt1 ∧ (nx ≠ [] ∧ m = nx.last ∧ lpt1[m].flags = LP_GROUP_NODES ∨ lpt1[m].flags = LP_INDEX_NODES) ↔ m < # lpt0 ∧ (nx ≠ [] ∧ m = nx.last ∧ lpt0[m].flags = LP_GROUP_NODES ∨ lpt0[m].flags = LP_INDEX_NODES), bleb = nobuffer

Rule argument

spec: bufleb name: elim seq: ⊦ bleb ≠ nobuffer → (m = bleb.leb ↔ bleb = buffered(m)) subst: [bleb, m] → [bleb, m]

Additional information

Used simplifier rules: ⊦ bleb ≠ nobuffer → (m = bleb.leb ↔ bleb = buffered(m)); (elim) Used terms: bleb, m