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Proof step 4 from proof bufs-log-only-id-idx (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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)

Rule argument

Additional information

Used simplifier rules: ⊦ nobuffer ≠ buffered(m); (s)