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Proof step 2 from proof bufs-log-only-id-idx (previous step)

id-idx(lpt0, lpt1) ⊦ bufs-log-only(bleb, nx, lpt1) ↔ bufs-log-only(bleb, nx, lpt0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
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

Rule argument

Additional information

Used simplifier rules: bufs-log-only: ⊦ bufs-log-only(bleb, nx, lpt) ↔ bleb = nobuffer ∨ bleb.leb < # lpt ∧ ( nx ≠ [] ∧ bleb.leb = nx.last ∧ lpt[bleb.leb].flags = LP_GROUP_NODES ∨ lpt[bleb.leb].flags = LP_INDEX_NODES);