Theorem ax11indn 2195

Description: Induction step for constructing a substitution instance of ax-11o 2141 without using ax-11o 2141. Negation case. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
ax11indn.1	|- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))

Assertion

Ref	Expression
ax11indn	|- (-. A.x x = y -> (x = y -> (-. ph -> A.x(x = y -> -. ph))))

Proof of Theorem ax11indn

Step	Hyp	Ref	Expression
1		19.8a 1756	. . 3 |- ((x = y /\ -. ph) -> E.x(x = y /\ -. ph))
2		exanali 1585	. . . 4 |- (E.x(x = y /\ -. ph) <-> -. A.x(x = y -> ph))
3		hbn1 1730	. . . . 5 |- (-. A.x x = y -> A.x -. A.x x = y)
4		hbn1 1730	. . . . 5 |- (-. A.x(x = y -> ph) -> A.x -. A.x(x = y -> ph))
5		ax11indn.1	. . . . . . 7 |- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))
6		con3 126	. . . . . . 7 |- ((ph -> A.x(x = y -> ph)) -> (-. A.x(x = y -> ph) -> -. ph))
7	5, 6	syl6 29	. . . . . 6 |- (-. A.x x = y -> (x = y -> (-. A.x(x = y -> ph) -> -. ph)))
8	7	com23 72	. . . . 5 |- (-. A.x x = y -> (-. A.x(x = y -> ph) -> (x = y -> -. ph)))
9	3, 4, 8	alrimdh 1587	. . . 4 |- (-. A.x x = y -> (-. A.x(x = y -> ph) -> A.x(x = y -> -. ph)))
10	2, 9	syl5bi 208	. . 3 |- (-. A.x x = y -> (E.x(x = y /\ -. ph) -> A.x(x = y -> -. ph)))
11	1, 10	syl5 28	. 2 |- (-. A.x x = y -> ((x = y /\ -. ph) -> A.x(x = y -> -. ph)))
12	11	exp3a 425	1 |- (-. A.x x = y -> (x = y -> (-. ph -> A.x(x = y -> -. ph))))

Syntax hints:  -. wn 3   -> wi 4   /\ wa 358  A.wal 1540  E.wex 1541

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746

This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542

This theorem is referenced by:  ax11indi  2196