Metamath Proof Explorer


Theorem bj-xpimasn

Description: The image of a singleton, general case. [Change and relabel xpimasn accordingly, maybe to xpima2sn.] (Contributed by BJ, 6-Apr-2019)

Ref Expression
Assertion bj-xpimasn ( ( 𝐴 × 𝐵 ) “ { 𝑋 } ) = if ( 𝑋𝐴 , 𝐵 , ∅ )

Proof

Step Hyp Ref Expression
1 xpima ( ( 𝐴 × 𝐵 ) “ { 𝑋 } ) = if ( ( 𝐴 ∩ { 𝑋 } ) = ∅ , ∅ , 𝐵 )
2 disjsn ( ( 𝐴 ∩ { 𝑋 } ) = ∅ ↔ ¬ 𝑋𝐴 )
3 eqid 𝐵 = 𝐵
4 2 3 ifbieq2i if ( ( 𝐴 ∩ { 𝑋 } ) = ∅ , ∅ , 𝐵 ) = if ( ¬ 𝑋𝐴 , ∅ , 𝐵 )
5 ifnot if ( ¬ 𝑋𝐴 , ∅ , 𝐵 ) = if ( 𝑋𝐴 , 𝐵 , ∅ )
6 1 4 5 3eqtri ( ( 𝐴 × 𝐵 ) “ { 𝑋 } ) = if ( 𝑋𝐴 , 𝐵 , ∅ )