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Proof step 2 from proof sub-restr-sub (previous step)

⊦ ((ns -- n) \ ns0) -- n = (ns \ ns0) -- n, (ns -- n) \ ns0 = ns \ ns0

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:
⊦ ((ns -- n) \ ns0) -- n = (ns \ ns0) -- n, ∀ n0. n0 ∈ (ns -- n) \ ns0 ↔ n0 ∈ ns \ ns0

Rule argument

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

Additional information

Used terms: ns \ ns0, (ns -- n) \ ns0