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Proof step 2 from proof bufs-log-only-write-lpt-same-flags (previous step)

lpt[n].flags = lp.flags, bufs-log-only(bleb, nx, lpt) ⊦ bufs-log-only(bleb, nx, lpt[n, lp])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
lpt[n].flags = lp.flags, bleb.leb < # lpt, nx ≠ [] ∧ bleb.leb = nx.last ∧ lpt[bleb.leb].flags = LP_GROUP_NODES ∨ lpt[bleb.leb].flags = LP_INDEX_NODES ⊦ lpt[n, lp][bleb.leb].flags = LP_INDEX_NODES, bleb = nobuffer, nx ≠ [] ∧ bleb.leb = nx.last ∧ lpt[n, lp][bleb.leb].flags = LP_GROUP_NODES

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);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # lpt[n, lp] = # lpt; (s)