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Proof step 17 from proof count-entries (previous step)

∀ str0. str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt0[str, tree], ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 = # sdt0, str ∈ sdt0, str ∈ entries0, ∀ str. str ∈ entries0 ↔ str ∈ sdt0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
∀ str0. str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt0[str, tree], ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 = # sdt0, str0 ∈ entries0 ↔ str0 ∈ sdt0, str ∈ entries0, str ∈ sdt0

Proof step 17 from proof count-entries (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information