Metamath Proof Explorer


Theorem 2nd1st

Description: Swap the members of an ordered pair. (Contributed by NM, 31-Dec-2014)

Ref Expression
Assertion 2nd1st A B × C A -1 = 2 nd A 1 st A

Proof

Step Hyp Ref Expression
1 1st2nd2 A B × C A = 1 st A 2 nd A
2 1 sneqd A B × C A = 1 st A 2 nd A
3 2 cnveqd A B × C A -1 = 1 st A 2 nd A -1
4 3 unieqd A B × C A -1 = 1 st A 2 nd A -1
5 opswap 1 st A 2 nd A -1 = 2 nd A 1 st A
6 4 5 eqtrdi A B × C A -1 = 2 nd A 1 st A