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Proof step 5 from proof entries-empty (previous step)

((abs(sdt, ∅) * D))(dirs) ⊦ sdt = ∅ ∧ D(dirs)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
sdt ≠ ∅ ⊦ ((abs(sdt, sdi) * D))(dirs) ↔ (∃ tree, ino, str, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs))
Proof step 7:
⊦ abs(∅, entries) = (⌜entries = ∅⌝ ∧ emp)

Rule argument

spec: <toplevel> name: entries-any seq: sdt ≠ ∅ ⊦ ((abs(sdt, sdi) * D))(dirs) ↔ (∃ tree, ino, str, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs)) subst: [dirs, D, sdi, sdt] → [dirs, D, ∅, sdt] flags: <all paths>

Additional information

Used simplifier rules: abs-empty: ⊦ abs(∅, entries) = (⌜entries = ∅⌝ ∧ emp);% ⊦ entries[str, ino] ≠ ∅; (s)% ⊦ ⌜true⌝ = true; (s)% ⊦ (true ∧ D) = D; (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]% ⊦ emp * D = D; (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] Used terms: dirs, D, ∅, sdt