Theorem ssopab2 4713

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 1788	. . . 4 ⊢ Ⅎx∀x∀y(φ → ψ)
2		nfa1 1788	. . . . . 6 ⊢ Ⅎy∀y(φ → ψ)
3		sp 1747	. . . . . . 7 ⊢ (∀y(φ → ψ) → (φ → ψ))
4	3	anim2d 548	. . . . . 6 ⊢ (∀y(φ → ψ) → ((z = ⟨x, y⟩ ∧ φ) → (z = ⟨x, y⟩ ∧ ψ)))
5	2, 4	eximd 1770	. . . . 5 ⊢ (∀y(φ → ψ) → (∃y(z = ⟨x, y⟩ ∧ φ) → ∃y(z = ⟨x, y⟩ ∧ ψ)))
6	5	sps 1754	. . . 4 ⊢ (∀x∀y(φ → ψ) → (∃y(z = ⟨x, y⟩ ∧ φ) → ∃y(z = ⟨x, y⟩ ∧ ψ)))
7	1, 6	eximd 1770	. . 3 ⊢ (∀x∀y(φ → ψ) → (∃x∃y(z = ⟨x, y⟩ ∧ φ) → ∃x∃y(z = ⟨x, y⟩ ∧ ψ)))
8	7	ss2abdv 3340	. 2 ⊢ (∀x∀y(φ → ψ) → {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ φ)} ⊆ {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ ψ)})
9		df-opab 4624	. 2 ⊢ {⟨x, y⟩ ∣ φ} = {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ φ)}
10		df-opab 4624	. 2 ⊢ {⟨x, y⟩ ∣ ψ} = {z ∣ ∃x∃y(z = ⟨x, y⟩ ∧ ψ)}
11	8, 9, 10	3sstr4g 3313	1 ⊢ (∀x∀y(φ → ψ) → {⟨x, y⟩ ∣ φ} ⊆ {⟨x, y⟩ ∣ ψ})

This proof depends on syntax axioms:   → wi 4   ∧ wa 358  ∀wal 1540  ∃wex 1541   = wceq 1642  {cab 2339   ⊆ wss 3258  ⟨cop 4562  {copab 4623

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260  df-opab 4624

This theorem is used by:  ssopab2b  4714  ssopab2i  4715  ssopab2dv  4716