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Proof step 15 from proof count-entries (previous step)

∀ str0. str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt, ∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0, ∀ 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 16:
∀ 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 ∈ entries0, str ∈ sdt0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ ¬ str ∈ sdt → # sdt[str, tree] = # sdt + 1; (s)