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Proof step 2 from proof map-set-empty (previous step)

⊦ # (∅).map(fn) = 0, (∅).map(fn) = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0)
Proof step 4:
⊦ n ∈ ns.map(fn) ↔ (∃ n0. n = fn(n0) ∧ n0 ∈ ns)

Rule argument

spec: set-basic, inst: nat-list-set name: Extension seq: ⊦ ns = ns0 ↔ (∀ n. n ∈ ns ↔ n ∈ ns0) subst: [ns0, ns] → [∅, (∅).map(fn)] flags: <all paths>

Additional information

Used simplifier rules: mapset-def: ⊦ n ∈ ns.map(fn) ↔ (∃ n0. n = fn(n0) ∧ n0 ∈ ns);% ⊦ ¬ n ∈ ∅; (s) Used terms: ∅, (∅).map(fn)