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Proof step 6 from proof entries-unfold (previous step)

((abs(sdt, sdi) * D))(dirs), str ∈ sdi, str ∈ sdt ⊦ ∃ tree, ino, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ str ∈ entries ↔ (∃ ino, entries0. entries = entries0[str, ino] ∧ ¬ str ∈ entries0)
Proof step 8:
((abs(sdt, sdi) * D))(dirs), str ∈ sdt, ∃ ino, entries0. sdi = entries0[str, ino] ∧ ¬ str ∈ entries0 ⊦ ∃ tree, ino, sdi0, sdt0. ¬ str ∈ sdt0 ∧ sdt = sdt0[str, tree] ∧ ¬ str ∈ sdi0 ∧ sdi = sdi0[str, ino] ∧ ((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs)

Rule argument

spec: store, inst: dir-entries name: exin seq: ⊦ str ∈ entries ↔ (∃ ino, entries0. entries = entries0[str, ino] ∧ ¬ str ∈ entries0) subst: [entries, str] → [sdi, str] flags: <all paths>

Additional information

Used terms: sdi, str