Theorem ssoprab2

Description: Equivalence of ordered pair abstraction subclass and implication. Compare ssopab2 . (Contributed by FL, 6-Nov-2013) (Proof shortened by Mario Carneiro, 11-Dec-2016)

	Ref	Expression
Assertion	ssoprab2	⊢ ( ∀ 𝑥 ∀ 𝑦 ∀ 𝑧 ( 𝜑 → 𝜓 ) → { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜑 } ⊆ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } )

Proof

Step	Hyp	Ref	Expression
1		id	⊢ ( ( 𝜑 → 𝜓 ) → ( 𝜑 → 𝜓 ) )
2	1	anim2d	⊢ ( ( 𝜑 → 𝜓 ) → ( ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜑 ) → ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ) )
3	2	aleximi	⊢ ( ∀ 𝑧 ( 𝜑 → 𝜓 ) → ( ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜑 ) → ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ) )
4	3	aleximi	⊢ ( ∀ 𝑦 ∀ 𝑧 ( 𝜑 → 𝜓 ) → ( ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜑 ) → ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ) )
5	4	aleximi	⊢ ( ∀ 𝑥 ∀ 𝑦 ∀ 𝑧 ( 𝜑 → 𝜓 ) → ( ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜑 ) → ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ) )
6	5	ss2abdv	⊢ ( ∀ 𝑥 ∀ 𝑦 ∀ 𝑧 ( 𝜑 → 𝜓 ) → { 𝑤 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜑 ) } ⊆ { 𝑤 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) } )
7		df-oprab	⊢ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜑 } = { 𝑤 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜑 ) }
8		df-oprab	⊢ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { 𝑤 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑤 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) }
9	6 7 8	3sstr4g	⊢ ( ∀ 𝑥 ∀ 𝑦 ∀ 𝑧 ( 𝜑 → 𝜓 ) → { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜑 } ⊆ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } )