Theorem csbopeq1a

Description: Equality theorem for substitution of a class A for an ordered pair <. x , y >. in B (analogue of csbeq1a ). (Contributed by NM, 19-Aug-2006) (Revised by Mario Carneiro, 31-Aug-2015)

	Ref	Expression
Assertion	csbopeq1a	$$⊢A=⟨x,y⟩\to ⦋1^{st}\left(A\right)/x⦌⦋2^{nd}\left(A\right)/y⦌B=B$$

Proof

Step	Hyp	Ref	Expression
1		vex	$$⊢x\in \mathrm{V}$$
2		vex	$$⊢y\in \mathrm{V}$$
3	1 2	op2ndd	$$⊢A=⟨x,y⟩\to 2^{nd}\left(A\right)=y$$
4	3	eqcomd	$$⊢A=⟨x,y⟩\to y=2^{nd}\left(A\right)$$
5		csbeq1a	$$⊢y=2^{nd}\left(A\right)\to B=⦋2^{nd}\left(A\right)/y⦌B$$
6	4 5	syl	$$⊢A=⟨x,y⟩\to B=⦋2^{nd}\left(A\right)/y⦌B$$
7	1 2	op1std	$$⊢A=⟨x,y⟩\to 1^{st}\left(A\right)=x$$
8	7	eqcomd	$$⊢A=⟨x,y⟩\to x=1^{st}\left(A\right)$$
9		csbeq1a	$$⊢x=1^{st}\left(A\right)\to ⦋2^{nd}\left(A\right)/y⦌B=⦋1^{st}\left(A\right)/x⦌⦋2^{nd}\left(A\right)/y⦌B$$
10	8 9	syl	$$⊢A=⟨x,y⟩\to ⦋2^{nd}\left(A\right)/y⦌B=⦋1^{st}\left(A\right)/x⦌⦋2^{nd}\left(A\right)/y⦌B$$
11	6 10	eqtr2d	$$⊢A=⟨x,y⟩\to ⦋1^{st}\left(A\right)/x⦌⦋2^{nd}\left(A\right)/y⦌B=B$$