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Proof step 4 from proof entries-any (previous step)

((abs(sdt, sdi) * D))(dirs) ⊦ sdt = ∅, ∃ 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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ sdt = ∅ ↔ ¬(∃ str, tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 6:
((abs(sdt, sdi) * D))(dirs) ⊦ ∀ str, tree, sdt0. ¬(sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0), ∃ 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)

Rule argument

spec: store, inst: directory-tree-inst name: exnotempty seq: ⊦ sdt = ∅ ↔ ¬(∃ str, tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0) subst: [sdt] → [sdt] flags: <all paths>

Additional information

Used terms: sdt