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Proof step 13 from proof nu-is-UN (previous step)

∀ x. compl(f)(S)(x) → S(x) ⊦ S(x), ∃ S. S ⊆ f(S) ∧ S(x)

User applied rule exists right

And obtained the following new goals

Proof step 14:
∀ x. compl(f)(S)(x) → S(x) ⊦ S(x), (λ x0. ¬ S(x0)) ⊆ f(λ x0. ¬ S(x0)) ∧ ¬ S(x)

Rule argument

pos: right 2 terms: λ x0. ¬ S(x0) abort: F discard: T computed: F

Additional information

Used terms: λ x0. ¬ S(x0)