Metamath Proof Explorer


Theorem ex-opab

Description: Example for df-opab . Example by David A. Wheeler. (Contributed by Mario Carneiro, 18-Jun-2015)

Ref Expression
Assertion ex-opab R = x y | x y x + 1 = y 3 R 4

Proof

Step Hyp Ref Expression
1 3cn 3
2 4cn 4
3 3p1e4 3 + 1 = 4
4 1 elexi 3 V
5 2 elexi 4 V
6 eleq1 x = 3 x 3
7 oveq1 x = 3 x + 1 = 3 + 1
8 7 eqeq1d x = 3 x + 1 = y 3 + 1 = y
9 6 8 3anbi13d x = 3 x y x + 1 = y 3 y 3 + 1 = y
10 eleq1 y = 4 y 4
11 eqeq2 y = 4 3 + 1 = y 3 + 1 = 4
12 10 11 3anbi23d y = 4 3 y 3 + 1 = y 3 4 3 + 1 = 4
13 eqid x y | x y x + 1 = y = x y | x y x + 1 = y
14 4 5 9 12 13 brab 3 x y | x y x + 1 = y 4 3 4 3 + 1 = 4
15 1 2 3 14 mpbir3an 3 x y | x y x + 1 = y 4
16 breq R = x y | x y x + 1 = y 3 R 4 3 x y | x y x + 1 = y 4
17 15 16 mpbiri R = x y | x y x + 1 = y 3 R 4