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Proof step 3 from proof abs-is-dnode (previous step)

tree.dir? ⊦ abs(tree, ino) = _∃(λ ad. ino => ad * abs(tree.entries, ad.entries) ∧ ⌜ad.meta = tree.meta⌝ ∧ ⌜ad.size = # ad.entries⌝ )

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: directory-tree name: dnode-elim seq: tree.dir? ⊦ md = tree.meta ∧ sdt = tree.entries ↔ tree = dnode(md, sdt) subst: [tree, md, sdt] → [tree, md, sdt]

Additional information

Used simplifier rules: abs-dnode: ⊦ abs(dnode(md, sdt), ino) = _∃ (λ ad. ino => ad * abs(sdt, ad.entries) ∧ ⌜ad.meta = md⌝ ∧ ⌜ad.size = # ad.entries⌝ );%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]%and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%tree.dir? ⊦ md = tree.meta ∧ sdt = tree.entries ↔ tree = dnode(md, sdt); (elim) Used terms: tree, md, sdt