Theorem sbcom3 2240

Description: Substituting y for x and then z for y is equivalent to substituting z for both x and y. (Contributed by Giovanni Mascellani, 8-Apr-2018.) Remove dependency on ax-11 1920. (Revised by Wolf Lammen, 16-Sep-2018.) (Proof shortened by Wolf Lammen, 16-Sep-2018.)

Assertion

Ref	Expression
sbcom3	|- ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph )

Proof of Theorem sbcom3

Step	Hyp	Ref	Expression
1		nfa1 1979	. . 3 |- F/ y A. y y = z
2		drsb2 2207	. . 3 |- ( A. y y = z -> ( [ y / x ] ph <-> [ z / x ] ph ) )
3	1, 2	sbbid 2232	. 2 |- ( A. y y = z -> ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph ) )
4		sb4b 2188	. . . 4 |- ( -. A. y y = z -> ( [ z / y ] [ y / x ] ph <-> A. y ( y = z -> [ y / x ] ph ) ) )
5		sbequ 2205	. . . . . 6 |- ( y = z -> ( [ y / x ] ph <-> [ z / x ] ph ) )
6	5	pm5.74i 249	. . . . 5 |- ( ( y = z -> [ y / x ] ph ) <-> ( y = z -> [ z / x ] ph ) )
7	6	albii 1691	. . . 4 |- ( A. y ( y = z -> [ y / x ] ph ) <-> A. y ( y = z -> [ z / x ] ph ) )
8	4, 7	syl6bb 265	. . 3 |- ( -. A. y y = z -> ( [ z / y ] [ y / x ] ph <-> A. y ( y = z -> [ z / x ] ph ) ) )
9		sb4b 2188	. . 3 |- ( -. A. y y = z -> ( [ z / y ] [ z / x ] ph <-> A. y ( y = z -> [ z / x ] ph ) ) )
10	8, 9	bitr4d 260	. 2 |- ( -. A. y y = z -> ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph ) )
11	3, 10	pm2.61i 168	1 |- ( [ z / y ] [ y / x ] ph <-> [ z / y ] [ z / x ] ph )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 188   A.wal 1442   [wsb 1797

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1669  ax-4 1682  ax-5 1758  ax-6 1805  ax-7 1851  ax-10 1915  ax-12 1933  ax-13 2091

This theorem depends on definitions:  df-bi 189  df-or 372  df-an 373  df-ex 1664  df-nf 1668  df-sb 1798

This theorem is referenced by:  sbco  2241  sbidm  2243  sbcom  2247  equsb3  2261  wl-equsb3  31884