Theorem ssopab2 4321

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		nfa1 1563	. . . 4 ⊢ Ⅎ𝑥∀𝑥∀𝑦(𝜑 → 𝜓)
2		nfa1 1563	. . . . . 6 ⊢ Ⅎ𝑦∀𝑦(𝜑 → 𝜓)
3		sp 1533	. . . . . . 7 ⊢ (∀𝑦(𝜑 → 𝜓) → (𝜑 → 𝜓))
4	3	anim2d 337	. . . . . 6 ⊢ (∀𝑦(𝜑 → 𝜓) → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → (𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
5	2, 4	eximd 1634	. . . . 5 ⊢ (∀𝑦(𝜑 → 𝜓) → (∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
6	5	sps 1559	. . . 4 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → (∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
7	1, 6	eximd 1634	. . 3 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
8	7	ss2abdv 3265	. 2 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ⊆ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)})
9		df-opab 4105	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)}
10		df-opab 4105	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
11	8, 9, 10	3sstr4g 3235	1 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → {⟨𝑥, 𝑦⟩ ∣ 𝜑} ⊆ {⟨𝑥, 𝑦⟩ ∣ 𝜓})

Syntax hints:   → wi 4   ∧ wa 104  ∀wal 1370   = wceq 1372  ∃wex 1514  {cab 2190   ⊆ wss 3165  ⟨cop 3635  {copab 4103

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-in 3171  df-ss 3178  df-opab 4105

This theorem is referenced by:  ssopab2b  4322  ssopab2i  4323  ssopab2dv  4324