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

lpt[n] = mklp(n0, n1, LP_FREE, n2), bufs-log-only(buffered(n), freelist, lpt) ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

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);% ⊦ LP_FREE ≠ LP_GROUP_NODES; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ LP_FREE ≠ LP_INDEX_NODES; (s)% ⊦ mklp(n2, n3, lpf, m).flags = lpf; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ buffered(m).leb = m; (s)% ⊦ nobuffer ≠ buffered(m); (s)