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Metamath Proof Explorer

Theorem ssopab2

Description: Equivalence of ordered pair abstraction subclass and implication. (Contributed by NM, 27-Dec-1996) (Revised by Mario Carneiro, 19-May-2013)

Ref Expression
Assertion ssopab2 ( ∀ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) → { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } ⊆ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜓 } )

Proof

Step Hyp Ref Expression
1 id ⊢ ( ( 𝜑 → 𝜓 ) → ( 𝜑 → 𝜓 ) )
2 1 anim2d ⊢ ( ( 𝜑 → 𝜓 ) → ( ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) → ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) ) )
3 2 aleximi ⊢ ( ∀ 𝑦 ( 𝜑 → 𝜓 ) → ( ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) → ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) ) )
4 3 aleximi ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) → ( ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) → ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) ) )
5 4 ss2abdv ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) → { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) } ⊆ { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) } )
6 df-opab ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } = { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) }
7 df-opab ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜓 } = { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) }
8 5 6 7 3sstr4g ⊢ ( ∀ 𝑥 ∀ 𝑦 ( 𝜑 → 𝜓 ) → { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } ⊆ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜓 } )