Theorem sbco2 913

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 877	. . . . 5 |- (x = y -> ([x / z][z / x]ph <-> [y / z][z / x]ph))
2		sbco2.1	. . . . . 6 |- (ph -> A.zph)
3	2	sbid2 911	. . . . 5 |- ([x / z][z / x]ph <-> ph)
4	1, 3	syl5bbr 412	. . . 4 |- (x = y -> (ph <-> [y / z][z / x]ph))
5		sbequ12 865	. . . 4 |- (x = y -> (ph <-> [y / x]ph))
6	4, 5	bitr3d 408	. . 3 |- (x = y -> ([y / z][z / x]ph <-> [y / x]ph))
7	6	a4s 682	. 2 |- (A.x x = y -> ([y / z][z / x]ph <-> [y / x]ph))
8		eq6 826	. . . 4 |- (-. A.x x = y -> A.x -. A.x x = y)
9	2	hbsb3 875	. . . . 5 |- ([z / x]ph -> A.x[z / x]ph)
10	9	hbsb4 905	. . . 4 |- (-. A.x x = y -> ([y / z][z / x]ph -> A.x[y / z][z / x]ph))
11	4	a1i 7	. . . 4 |- (-. A.x x = y -> (x = y -> (ph <-> [y / z][z / x]ph)))
12	8, 10, 11	sbied 903	. . 3 |- (-. A.x x = y -> ([y / x]ph <-> [y / z][z / x]ph))
13	12	bicomd 399	. 2 |- (-. A.x x = y -> ([y / z][z / x]ph <-> [y / x]ph))
14	7, 13	pm2.61i 110	1 |- ([y / z][z / x]ph <-> [y / x]ph)

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127  A.wal 672   = weq 797  [wsb 852

This theorem is referenced by:  sbco2d 914  sb7 991  sbralie 1439  sbcco 1448

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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