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Proof step 32 from proof dir-empty (previous step)

tree[p] = dnode(md, sdt), sdt[str] = dnode(md0, sdt0), (( entries[str] => mkdir(# entries0, md0, # links(entries[str], dirs), m2, entries0) * abs(sdt0, entries0) * m => mkdir(# entries, md, # links(m, dirs), n1, entries) * abs(sdt -- str, entries -- str) * abs|(tree, ino, p, m)) )(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), str ∈ entries, p ∈ tree, str ∈ sdt ⊦ entries0 = ∅ ↔ sdt0 = ∅

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ abs(∅, entries) = (⌜entries = ∅⌝ ∧ emp); (s)% ⊦ ⌜false⌝ = false; (s)% ⊦ (false ∧ D) = false; (s)%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]% ⊦ false * D = false; (s)%sep-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]% ⊦ ¬ false(dirs); (s)% ⊦ # ∅ = 0; (s)% ⊦ abs(sdt, ∅) = (⌜sdt = ∅⌝ ∧ emp); (s)