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Proof step 17 from proof in-below (previous step)

n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m, n0 < m, n = n0 ∨ n ∈ ns0 ∧ n < m, ∀ ns1, n, m. # ns1 < #(ns0 ++ n0) → true → (n ∈ below(ns1, m) ↔ n ∈ ns1 ∧ n < m) ⊦ #(below(ns0, m) ++ n0) = 0, n0 ∈ ns0, (n = n0 ∨ n ∈ ns0) ∧ n < m

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ ¬ n ∈ ns → #(ns ++ n) = # ns + 1; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic