Theorem sbcco3g 2031

Description: Composition of two substitutions.

Hypothesis

Ref	Expression
sbcco3g.1	|- (x = A -> B = C)

Assertion

Ref	Expression
sbcco3g	|- ((A e. R /\ A.x B e. S) -> ([A / x][B / y]ph <-> [C / y]ph))

Distinct variable groups:   x,A   ph,x   x,C   x,y

Proof of Theorem sbcco3g

Step	Hyp	Ref	Expression
1		sbcnestg 2028	. 2 |- ((A e. R /\ A.x B e. S) -> ([A / x][B / y]ph <-> [[_A / x]_B / y]ph))
2		ax-17 968	. . . . . 6 |- (z e. C -> A.x z e. C)
3	2	gen2 980	. . . . 5 |- A.xA.z(z e. C -> A.x z e. C)
4		sbcco3g.1	. . . . . 6 |- (x = A -> B = C)
5	4	ax-gen 960	. . . . 5 |- A.x(x = A -> B = C)
6		csbiegft 2019	. . . . 5 |- ((A e. R /\ A.xA.z(z e. C -> A.x z e. C) /\ A.x(x = A -> B = C)) -> [_A / x]_B = C)
7	3, 5, 6	mp3an23 905	. . . 4 |- (A e. R -> [_A / x]_B = C)
8		dfsbcq 1933	. . . 4 |- ([_A / x]_B = C -> ([[_A / x]_B / y]ph <-> [C / y]ph))
9	7, 8	syl 10	. . 3 |- (A e. R -> ([[_A / x]_B / y]ph <-> [C / y]ph))
10	9	adantr 389	. 2 |- ((A e. R /\ A.x B e. S) -> ([[_A / x]_B / y]ph <-> [C / y]ph))
11	1, 10	bitrd 526	1 |- ((A e. R /\ A.x B e. S) -> ([A / x][B / y]ph <-> [C / y]ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 951   = wceq 953   e. wcel 955  [wsbc 1166  [_csb 1991

This theorem is referenced by:  fzshftralt 6454

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-9 962  ax-10 963  ax-11 964  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-16 1206  ax-11o 1213  ax-ext 1452

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 775  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-v 1803  df-sbc 1932  df-csb 1992

Copyright terms: Public domain