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

str ∈ entries0[str, ino] ↔ str ∈ sdt, ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
str ∈ sdt, ∀ 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[str0, ino] ↔ str = str0 ∨ str ∈ entries; (ws)% ⊦ str ∈ entries[str, ino]; (s)