Theorem fconstmpt 5681

Description: Representation of a constant function using the mapping operation. (Note that x cannot appear free in B.) (Contributed by set.mm contributors, 16-Nov-2013.)

Assertion

Ref	Expression
fconstmpt	⊢ (A × {B}) = (x ∈ A ↦ B)

Distinct variable groups:   x,A   x,B

Proof of Theorem fconstmpt

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		fconstopab 4815	. 2 ⊢ (A × {B}) = {⟨x, y⟩ ∣ (x ∈ A ∧ y = B)}
2		df-mpt 5652	. 2 ⊢ (x ∈ A ↦ B) = {⟨x, y⟩ ∣ (x ∈ A ∧ y = B)}
3	1, 2	eqtr4i 2376	1 ⊢ (A × {B}) = (x ∈ A ↦ B)

Syntax hints:   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {csn 3737  {copab 4622   × cxp 4770   ↦ cmpt 5651

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  df-mpt 5652

This theorem is referenced by: (None)