Metamath Proof Explorer


Theorem axc9

Description: Derive set.mm's original ax-c9 from the shorter ax-13 . Usage is discouraged to avoid uninformed ax-13 propagation. (Contributed by NM, 29-Nov-2015) (Revised by NM, 24-Dec-2015) (Proof shortened by Wolf Lammen, 29-Apr-2018) (New usage is discouraged.)

Ref Expression
Assertion axc9 ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( ¬ ∀ 𝑧 𝑧 = 𝑦 → ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) )

Proof

Step Hyp Ref Expression
1 nfeqf ( ( ¬ ∀ 𝑧 𝑧 = 𝑥 ∧ ¬ ∀ 𝑧 𝑧 = 𝑦 ) → Ⅎ 𝑧 𝑥 = 𝑦 )
2 1 nf5rd ( ( ¬ ∀ 𝑧 𝑧 = 𝑥 ∧ ¬ ∀ 𝑧 𝑧 = 𝑦 ) → ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) )
3 2 ex ( ¬ ∀ 𝑧 𝑧 = 𝑥 → ( ¬ ∀ 𝑧 𝑧 = 𝑦 → ( 𝑥 = 𝑦 → ∀ 𝑧 𝑥 = 𝑦 ) ) )