Theorem sbcco3g 2586

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 2583	. 2 |- ((A e. R /\ A.x B e. S) -> ([A / x][B / y]ph <-> [[_A / x]_B / y]ph))
2		ax-17 1317	. . . . . 6 |- (z e. C -> A.x z e. C)
3	2	gen2 1329	. . . . 5 |- A.xA.z(z e. C -> A.x z e. C)
4		sbcco3g.1	. . . . . 6 |- (x = A -> B = C)
5	4	ax-gen 1305	. . . . 5 |- A.x(x = A -> B = C)
6		csbiegft 2573	. . . . 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 1183	. . . 4 |- (A e. R -> [_A / x]_B = C)
8		dfsbcq 2455	. . . 4 |- ([_A / x]_B = C -> ([[_A / x]_B / y]ph <-> [C / y]ph))
9	7, 8	syl 12	. . 3 |- (A e. R -> ([[_A / x]_B / y]ph <-> [C / y]ph))
10	9	adantr 425	. 2 |- ((A e. R /\ A.x B e. S) -> ([[_A / x]_B / y]ph <-> [C / y]ph))
11	1, 10	bitrd 587	1 |- ((A e. R /\ A.x B e. S) -> ([A / x][B / y]ph <-> [C / y]ph))

Syntax hints:   -> wi 3   <-> wb 163   /\ wa 240  A.wal 1296   = wceq 1298   e. wcel 1300  [wsbc 1534  [_csb 2540

This theorem is referenced by:  fzshftral 7701

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-5 1302  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-3an 860  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294  df-sbc 2454  df-csb 2541

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