Theorem ssopab2 5528

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

Assertion

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

Proof of Theorem ssopab2

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		id 23	. . . . . 6 ⊢ ((𝜑 → 𝜓) → (𝜑 → 𝜓))
2	1	anim2d 624	. . . . 5 ⊢ ((𝜑 → 𝜓) → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → (𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
3	2	aleximi 1865	. . . 4 ⊢ (∀𝑦(𝜑 → 𝜓) → (∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
4	3	aleximi 1865	. . 3 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
5	4	ss2abdv 3993	. 2 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ⊆ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)})
6		df-opab 5174	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)}
7		df-opab 5174	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
8	5, 6, 7	3sstr4g 3966	1 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → {⟨𝑥, 𝑦⟩ ∣ 𝜑} ⊆ {⟨𝑥, 𝑦⟩ ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wal 1568   = wceq 1570  ∃wex 1812  {cab 2741   ⊆ wss 3900  ⟨cop 4840  {copab 5173

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-ss 3901  df-opab 5174

This theorem is used by:  ssopab2bw  5529  ssopab2b  5531  ssopab2i  5532  ssopab2dv  5533  opabbrex  7504