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Proof step 2 from proof abs-is-file (previous step)

⊦ ((abs(tree, ino) * D))(dirs) ↔ tree = fnode(⌜ino⌝ ) ∧ D(dirs), ino ∈ dirs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ abs(fnode(fid), ino) = (emp ∧ ⌜ino = fid.ino⌝ )
Proof step 4:
((abs(tree, ino) * D))(dirs) ⊦ ino ∈ dirs, tree = fnode(⌜ino⌝ ) ∧ D(dirs)

Rule argument

Additional information

Used simplifier rules: abs-fnode: ⊦ abs(fnode(fid), ino) = (emp ∧ ⌜ino = fid.ino⌝ );% ⊦ true(dirs); (s)% ⊦ ⌜true⌝ = true; (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)% ⊦ ⌜m⌝ .ino = m; (s)