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Proof step 10 from proof keys-inodekey-in (previous step)

∃ ns0. ns = ns0 ++ n0 ∧ ¬ n0 ∈ ns0, ∀ ns0, n. # ns0 < # ns → true → (inodekey(n) ∈ ns0.keys ↔ n ∈ ns0) ⊦ # ns = 0, inodekey(n) ∈ ns.keys ↔ n ∈ ns

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ ns1, n. # ns1 < #(ns0 ++ n0) → true → (inodekey(n) ∈ ns1.keys ↔ n ∈ ns1) ⊦ inodekey(n) ∈ ns0.keys ↔ n ∈ ns0, n = n0, n0 ∈ ns0

Rule argument

Additional information

Used simplifier rules: keys-rec: ⊦ ¬ n ∈ ns → (ns ++ n).keys = ns.keys ++ inodekey(n);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ ¬ n ∈ ns → #(ns ++ n) = # ns + 1; (s)% ⊦ n ∈ ns ++ n0 ↔ n = n0 ∨ n ∈ ns; (s)% ⊦ inodekey(m) = inodekey(m0) ↔ m = m0; (s)% ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)