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Proof step 16 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 ∈ entries0, str ∈ sdt0

User applied rule apply induction

And obtained the following new goals

Proof step 17:
∀ 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

Rule argument

subst: [entries1, sdt] → [entries0, sdt0]

Additional information

Used simplifier rules: ⊦ ¬ str ∈ entries → # entries[str, ino] = # entries + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: entries0, sdt0