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Proof step 2 from proof abs-keys-empty (previous step)

⊦ # abs-keys(∅) = 0, abs-keys(∅) = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ns = ns0 ↔ (∀ ino. ino ∈ ns ↔ ino ∈ ns0)

Rule argument

spec: set-basic, inst: dir-heap-refset name: Extension seq: ⊦ ns = ns0 ↔ (∀ ino. ino ∈ ns ↔ ino ∈ ns0) subst: [ns0, ns] → [∅, abs-keys(∅)] flags: <all paths>

Additional information

Used simplifier rules: abs-keys: ⊦ n ∈ abs-keys(ks) ↔ inodekey(n) ∈ ks;% ⊦ ¬ key ∈ ∅; (s)% ⊦ ¬ ino ∈ ∅; (s) Used terms: ∅, abs-keys(∅)