Theorem fconstopab 4815

Description: Representation of a constant function using ordered pairs. (Contributed by NM, 12-Oct-1999.)

Assertion

Ref	Expression
fconstopab	⊢ (A × {B}) = {⟨x, y⟩ ∣ (x ∈ A ∧ y = B)}

Distinct variable groups:   x,y,A   x,B,y

Proof of Theorem fconstopab

Step	Hyp	Ref	Expression
1		df-xp 4784	. 2 ⊢ (A × {B}) = {⟨x, y⟩ ∣ (x ∈ A ∧ y ∈ {B})}
2		df-sn 3741	. . . . 5 ⊢ {B} = {y ∣ y = B}
3	2	abeq2i 2460	. . . 4 ⊢ (y ∈ {B} ↔ y = B)
4	3	anbi2i 675	. . 3 ⊢ ((x ∈ A ∧ y ∈ {B}) ↔ (x ∈ A ∧ y = B))
5	4	opabbii 4626	. 2 ⊢ {⟨x, y⟩ ∣ (x ∈ A ∧ y ∈ {B})} = {⟨x, y⟩ ∣ (x ∈ A ∧ y = B)}
6	1, 5	eqtri 2373	1 ⊢ (A × {B}) = {⟨x, y⟩ ∣ (x ∈ A ∧ y = B)}

Syntax hints:   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {csn 3737  {copab 4622   × cxp 4770

This theorem was proved from 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 theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-sn 3741  df-opab 4623  df-xp 4784

This theorem is referenced by:  fconst  5250  fopabsn  5441  fconstmpt  5681