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Proof step 19 from proof abs-tree (previous step)

tree0 = fnode(fid), ((abs(tree0, ino) * D0))(dirs), ((abs(tree1, ino) * D1))(dirs), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ (tree0.dir? ↔ tree1.dir?) ∧ (tree0.dir? ⊃ tree0.meta = tree1.meta;tree0.fid = tree1.fid)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
⊦ abs(fnode(fid), ino) = (emp ∧ ⌜ino = fid.ino⌝ )
Proof step 21:
((abs(tree1, fid.ino) * D1))(dirs), D0(dirs), ∀ ino0. ⌜ino0⌝ = fid → ino0 ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ ¬ tree1.dir? ∧ fid = tree1.fid

Rule argument

Additional information

Used simplifier rules: abs-fnode: ⊦ abs(fnode(fid), ino) = (emp ∧ ⌜ino = fid.ino⌝ );% ⊦ fnode(fid).fid = fid; (s)% ⊦ ¬ fnode(fid).dir?; (s)% ⊦ fid ∈ {fid0} ↔ fid = fid0; (s)% ⊦ fnode(fid).fids = {fid} ; (s)% ⊦ ⌜bv⌝ (dirs) ↔ bv; (s)% ⊦ ((D ∧ D0))(dirs) ↔ D(dirs) ∧ D0(dirs); (s)%and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]%sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]% ⊦ emp * D = D; (s)% ⊦ (⌜bv⌝ ∧ D) * D0 = (⌜bv⌝ ∧ D * D0); (s)