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Proof step 25 from proof abs-acyclic (previous step)

entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (⌜str ∈ entries⌝ ∧ abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str]))) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
tree[p] = dnode(md0, sdt), entries[str] = ROOT_INO, (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ sdt, str ∈ entries ⊦

Rule argument

Additional information

Used simplifier rules: 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]% ⊦ (true ∧ D) = D; (s)% ⊦ ⌜true⌝ = true; (s)