Theorem sbcco3g 2012

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 2009	. 2 |- ((A e. R /\ A.x B e. S) -> ([A / x][B / y]ph <-> [[_A / x]_B / y]ph))
2		ax-17 1190	. . . . . 6 |- (z e. C -> A.x z e. C)
3	2	gen2 959	. . . . 5 |- A.xA.z(z e. C -> A.x z e. C)
4		sbcco3g.1	. . . . . 6 |- (x = A -> B = C)
5	4	ax-gen 955	. . . . 5 |- A.x(x = A -> B = C)
6		csbiegft 2000	. . . . 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 904	. . . 4 |- (A e. R -> [_A / x]_B = C)
8		dfsbcq 1914	. . . 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 950   = wceq 1099   e. wcel 1105  [wsbc 1153  [_csb 1972

This theorem is referenced by:  fzshftralt 6405

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-7 954  ax-gen 955  ax-8 1101  ax-9 1102  ax-10 1103  ax-12 1104  ax-11 1180  ax-17 1190  ax-16 1194  ax-11o 1202  ax-ext 1436

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 774  df-ex 957  df-sb 1155  df-clab 1441  df-cleq 1446  df-clel 1449  df-v 1787  df-sbc 1913  df-csb 1973

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