Theorem sbco3 1257

Description: A composition law for substitution.

Assertion

Ref	Expression
sbco3	|- ([z / y][y / x]ph <-> [z / x][x / y]ph)

Proof of Theorem sbco3

Step	Hyp	Ref	Expression
1		drsb1 1175	. . 3 |- (A.x x = y -> ([z / x][y / x]ph <-> [z / y][y / x]ph))
2		sbequ12a 1183	. . . . 5 |- (x = y -> ([y / x]ph <-> [x / y]ph))
3	2	19.20i 992	. . . 4 |- (A.x x = y -> A.x([y / x]ph <-> [x / y]ph))
4		a4sbbi 1245	. . . 4 |- (A.x([y / x]ph <-> [x / y]ph) -> ([z / x][y / x]ph <-> [z / x][x / y]ph))
5	3, 4	syl 10	. . 3 |- (A.x x = y -> ([z / x][y / x]ph <-> [z / x][x / y]ph))
6	1, 5	bitr3d 530	. 2 |- (A.x x = y -> ([z / y][y / x]ph <-> [z / x][x / y]ph))
7		hbnae 1147	. . . 4 |- (-. A.x x = y -> A.y -. A.x x = y)
8		hbnae 1147	. . . 4 |- (-. A.x x = y -> A.x -. A.x x = y)
9		hbsb2 1227	. . . 4 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
10	7, 8, 9	sbco2d 1256	. . 3 |- (-. A.x x = y -> ([z / x][x / y][y / x]ph <-> [z / y][y / x]ph))
11		sbco 1252	. . . 4 |- ([x / y][y / x]ph <-> [x / y]ph)
12	11	sbbii 1174	. . 3 |- ([z / x][x / y][y / x]ph <-> [z / x][x / y]ph)
13	10, 12	syl5rbbr 535	. 2 |- (-. A.x x = y -> ([z / y][y / x]ph <-> [z / x][x / y]ph))
14	6, 13	pm2.61i 126	1 |- ([z / y][y / x]ph <-> [z / x][x / y]ph)

Syntax hints:  -. wn 2   <-> wb 146  A.wal 954  [wsbc 1170

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-9 965  ax-10 966  ax-11 967  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172

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