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Proof step 12 from proof entries-dom (previous step)

((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 = str0 ∨ str ∈ sdi0 ↔ str = str0 ∨ str ∈ sdt0, str0 ∈ sdi0, str0 ∈ sdt0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
((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 ↔ str ∈ sdt0, str = str0, str0 ∈ sdt0, str0 ∈ sdi0

Proof step 12 from proof entries-dom (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information