Theorem sbco3 915

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		del43b 857	. . 3 |- (A.x x = y -> ([z / x][y / x]ph <-> [z / y][y / x]ph))
2		sbequ12a 867	. . . . 5 |- (x = y -> ([y / x]ph <-> [x / y]ph))
3	2	19.20i 691	. . . 4 |- (A.x x = y -> A.x([y / x]ph <-> [x / y]ph))
4		sbba4 896	. . . 4 |- (A.x([y / x]ph <-> [x / y]ph) -> ([z / x][y / x]ph <-> [z / x][x / y]ph))
5	3, 4	syl 12	. . 3 |- (A.x x = y -> ([z / x][y / x]ph <-> [z / x][x / y]ph))
6	1, 5	bitr3d 408	. 2 |- (A.x x = y -> ([z / y][y / x]ph <-> [z / x][x / y]ph))
7		eq6 826	. . . 4 |- (-. A.x x = y -> A.y -. A.x x = y)
8		eq6 826	. . . 4 |- (-. A.x x = y -> A.x -. A.x x = y)
9		hbsb2 873	. . . 4 |- (-. A.x x = y -> ([y / x]ph -> A.x[y / x]ph))
10	7, 8, 9	sbco2d 914	. . 3 |- (-. A.x x = y -> ([z / x][x / y][y / x]ph <-> [z / y][y / x]ph))
11		sbco 910	. . . 4 |- ([x / y][y / x]ph <-> [x / y]ph)
12	11	bisb 855	. . 3 |- ([z / x][x / y][y / x]ph <-> [z / x][x / y]ph)
13	10, 12	syl5rbbr 413	. 2 |- (-. A.x x = y -> ([z / y][y / x]ph <-> [z / x][x / y]ph))
14	6, 13	pm2.61i 110	1 |- ([z / y][y / x]ph <-> [z / x][x / y]ph)

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

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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