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Proof step 11 from proof keys-inodekey-in (previous step)

∀ ns1, n. # ns1 < #(ns0 ++ n0) → true → (inodekey(n) ∈ ns1.keys ↔ n ∈ ns1) ⊦ inodekey(n) ∈ ns0.keys ↔ n ∈ ns0, n = n0, n0 ∈ ns0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [ns1, n] → [ns0, n]

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