Metamath Proof Explorer


Theorem csbnestgf

Description: Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker csbnestgfw when possible. (Contributed by NM, 23-Nov-2005) (Proof shortened by Mario Carneiro, 10-Nov-2016) (New usage is discouraged.)

Ref Expression
Assertion csbnestgf A V y _ x C A / x B / y C = A / x B / y C

Proof

Step Hyp Ref Expression
1 elex A V A V
2 df-csb B / y C = z | [˙B / y]˙ z C
3 2 abeq2i z B / y C [˙B / y]˙ z C
4 3 sbcbii [˙A / x]˙ z B / y C [˙A / x]˙ [˙B / y]˙ z C
5 nfcr _ x C x z C
6 5 alimi y _ x C y x z C
7 sbcnestgf A V y x z C [˙A / x]˙ [˙B / y]˙ z C [˙ A / x B / y]˙ z C
8 6 7 sylan2 A V y _ x C [˙A / x]˙ [˙B / y]˙ z C [˙ A / x B / y]˙ z C
9 4 8 syl5bb A V y _ x C [˙A / x]˙ z B / y C [˙ A / x B / y]˙ z C
10 9 abbidv A V y _ x C z | [˙A / x]˙ z B / y C = z | [˙ A / x B / y]˙ z C
11 1 10 sylan A V y _ x C z | [˙A / x]˙ z B / y C = z | [˙ A / x B / y]˙ z C
12 df-csb A / x B / y C = z | [˙A / x]˙ z B / y C
13 df-csb A / x B / y C = z | [˙ A / x B / y]˙ z C
14 11 12 13 3eqtr4g A V y _ x C A / x B / y C = A / x B / y C