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Proof step 7 from proof entries-dom (previous step)

∃ 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), ∀ sdt0, sdi, D, dirs, str. # sdt0 < # sdt → ((abs(sdt0, sdi) * D))(dirs) → (str ∈ sdi ↔ str ∈ sdt0) ⊦ str ∈ sdi ↔ str ∈ sdt, sdt = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
((abs(tree, ino) * abs(sdt0, sdi0) * D))(dirs), ∀ sdt1, sdi, D, dirs, str. # sdt1 < # sdt0[str0, tree] → ((abs(sdt1, sdi) * D))(dirs) → (str ∈ sdi ↔ str ∈ sdt1) ⊦ str ∈ sdi0[str0, ino] ↔ str ∈ sdt0[str0, tree], str0 ∈ sdi0, str0 ∈ sdt0

Rule argument

Additional information

Used simplifier rules: ⊦ sdt[str, tree] ≠ ∅; (s)