Theorem del43 856

Description: A distinctor elimination lemma for substitution.

Assertion

Ref	Expression
del43	|- (A.x x = y -> ([z / x]ph -> [z / y]ph))

Proof of Theorem del43

Step	Hyp	Ref	Expression
1		ax-8 798	. . . . . 6 |- (y = x -> (y = z -> x = z))
2	1	a4s 682	. . . . 5 |- (A.y y = x -> (y = z -> x = z))
3	2	eq4s 822	. . . 4 |- (A.x x = y -> (y = z -> x = z))
4	3	syl4d 28	. . 3 |- (A.x x = y -> ((x = z -> ph) -> (y = z -> ph)))
5		ax-8 798	. . . . . 6 |- (x = y -> (x = z -> y = z))
6	5	a4s 682	. . . . 5 |- (A.x x = y -> (x = z -> y = z))
7	6	anim1d 432	. . . 4 |- (A.x x = y -> ((x = z /\ ph) -> (y = z /\ ph)))
8	7	del40 839	. . 3 |- (A.x x = y -> (E.x(x = z /\ ph) -> E.y(y = z /\ ph)))
9	4, 8	anim12d 431	. 2 |- (A.x x = y -> (((x = z -> ph) /\ E.x(x = z /\ ph)) -> ((y = z -> ph) /\ E.y(y = z /\ ph))))
10		df-sb 853	. 2 |- ([z / x]ph <-> ((x = z -> ph) /\ E.x(x = z /\ ph)))
11		df-sb 853	. 2 |- ([z / y]ph <-> ((y = z -> ph) /\ E.y(y = z /\ ph)))
12	9, 10, 11	3imtr4g 426	1 |- (A.x x = y -> ([z / x]ph -> [z / y]ph))

Syntax hints:   -> wi 2   /\ wa 196  A.wal 672  E.wex 678   = weq 797  [wsb 852

This theorem is referenced by:  del43b 857  sbequi 876  sb9i 920

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-12 802

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

metamath.org