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Proof step 12 from proof in-below (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 13:
n ∈ below(ns0, m) ↔ n ∈ ns0 ∧ n < m, ∀ ns1, n, m. # ns1 < #(ns0 ++ n0) → true → (n ∈ below(ns1, m) ↔ n ∈ ns1 ∧ n < m) ⊦ n ∈ (n0 < m ⊃ below(ns0, m) ++ n0;below(ns0, m)) ↔ (n = n0 ∨ n ∈ ns0) ∧ n < m, n0 ∈ ns0

Rule argument

subst: [ns1, n, m] → [ns0, n, m]

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ns → #(ns ++ n) = # ns + 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: ns0, n, m