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Proof step 11 from proof keys-inodekeys (previous step)

∃ ns0. ns = ns0 ++ n ∧ ¬ n ∈ ns0, ∀ ns0. # ns0 < # ns → true → ns0.keys.inodes? ⊦ # ns = 0, ns.keys.inodes?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ ns1. # ns1 < #(ns0 ++ n) → true → ns1.keys.inodes? ⊦ (ns0.keys ++ inodekey(n)).inodes?, n ∈ 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)