Theorem sbco2 1250

Description: A composition law for substitution.

Hypothesis

Ref	Expression
sbco2.1	|- (ph -> A.zph)

Assertion

Ref	Expression
sbco2	|- ([y / z][z / x]ph <-> [y / x]ph)

Proof of Theorem sbco2

Step	Hyp	Ref	Expression
1		sbequ 1224	. . . . 5 |- (x = y -> ([x / z][z / x]ph <-> [y / z][z / x]ph))
2		sbco2.1	. . . . . 6 |- (ph -> A.zph)
3	2	sbid2 1248	. . . . 5 |- ([x / z][z / x]ph <-> ph)
4	1, 3	syl5bbr 532	. . . 4 |- (x = y -> (ph <-> [y / z][z / x]ph))
5		sbequ12 1177	. . . 4 |- (x = y -> (ph <-> [y / x]ph))
6	4, 5	bitr3d 528	. . 3 |- (x = y -> ([y / z][z / x]ph <-> [y / x]ph))
7	6	a4s 981	. 2 |- (A.x x = y -> ([y / z][z / x]ph <-> [y / x]ph))
8		hbnae 1143	. . . 4 |- (-. A.x x = y -> A.x -. A.x x = y)
9	2	hbsb3 1202	. . . . 5 |- ([z / x]ph -> A.x[z / x]ph)
10	9	hbsb4 1243	. . . 4 |- (-. A.x x = y -> ([y / z][z / x]ph -> A.x[y / z][z / x]ph))
11	4	a1i 8	. . . 4 |- (-. A.x x = y -> (x = y -> (ph <-> [y / z][z / x]ph)))
12	8, 10, 11	sbied 1191	. . 3 |- (-. A.x x = y -> ([y / x]ph <-> [y / z][z / x]ph))
13	12	bicomd 519	. 2 |- (-. A.x x = y -> ([y / z][z / x]ph <-> [y / x]ph))
14	7, 13	pm2.61i 126	1 |- ([y / z][z / x]ph <-> [y / x]ph)

Syntax hints:  -. wn 2   -> wi 3   <-> wb 146  A.wal 951  [wsbc 1166

This theorem is referenced by:  sbco2d 1251  equsb3 1325  elsb3 1326  dfsb7 1335  2eu6 1447  sbralie 1931  sbccog 1942

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-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-11o 1213

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 978  df-sb 1168

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