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Proof step 10 from proof count-entries (previous step)

∀ str. str ∈ entries ↔ str ∈ sdt, ∀ entries0, sdt. # entries0 < # entries → (∀ str. str ∈ entries0 ↔ str ∈ sdt) → # entries0 = # sdt ⊦ # entries = # sdt, ∀ str, ino, entries0. ¬(entries = entries0[str, ino] ∧ ¬ str ∈ entries0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ str0. str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt, ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 + 1 = # sdt, str ∈ entries0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ str ∈ entries → # entries[str, ino] = # entries + 1; (s)