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Proof step 4 from proof choice-aux (previous step)

ok(a), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ ∃ nf. nf(0) = a

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: λ n. a abort: F discard: T computed: F

Additional information

Used terms: λ n. a