Theorem ssopab2 4260

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 1534	. . . 4 ⊢ Ⅎ𝑥∀𝑥∀𝑦(𝜑 → 𝜓)
2		nfa1 1534	. . . . . 6 ⊢ Ⅎ𝑦∀𝑦(𝜑 → 𝜓)
3		sp 1504	. . . . . . 7 ⊢ (∀𝑦(𝜑 → 𝜓) → (𝜑 → 𝜓))
4	3	anim2d 335	. . . . . 6 ⊢ (∀𝑦(𝜑 → 𝜓) → ((𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → (𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
5	2, 4	eximd 1605	. . . . 5 ⊢ (∀𝑦(𝜑 → 𝜓) → (∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
6	5	sps 1530	. . . 4 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → (∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
7	1, 6	eximd 1605	. . 3 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → (∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑) → ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)))
8	7	ss2abdv 3220	. 2 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)} ⊆ {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)})
9		df-opab 4051	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜑} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜑)}
10		df-opab 4051	. 2 ⊢ {⟨𝑥, 𝑦⟩ ∣ 𝜓} = {𝑧 ∣ ∃𝑥∃𝑦(𝑧 = ⟨𝑥, 𝑦⟩ ∧ 𝜓)}
11	8, 9, 10	3sstr4g 3190	1 ⊢ (∀𝑥∀𝑦(𝜑 → 𝜓) → {⟨𝑥, 𝑦⟩ ∣ 𝜑} ⊆ {⟨𝑥, 𝑦⟩ ∣ 𝜓})

Syntax hints:   → wi 4   ∧ wa 103  ∀wal 1346   = wceq 1348  ∃wex 1485  {cab 2156   ⊆ wss 3121  ⟨cop 3586  {copab 4049

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-in 3127  df-ss 3134  df-opab 4051

This theorem is referenced by:  ssopab2b  4261  ssopab2i  4262  ssopab2dv  4263