Theorem ax11indn 1366

Description: Induction step for constructing a substitution instance of ax-11o 1218 without using ax-11o 1218. Negation case.

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		hbn1 1015	. . . . 5 |- (-. A.x x = y -> A.x -. A.x x = y)
2		hbn1 1015	. . . . 5 |- (-. A.x(x = y -> ph) -> A.x -. A.x(x = y -> ph))
3		ax11indn.1	. . . . . . 7 |- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))
4		con3 94	. . . . . . 7 |- ((ph -> A.x(x = y -> ph)) -> (-. A.x(x = y -> ph) -> -. ph))
5	3, 4	syl6 22	. . . . . 6 |- (-. A.x x = y -> (x = y -> (-. A.x(x = y -> ph) -> -. ph)))
6	5	com23 32	. . . . 5 |- (-. A.x x = y -> (-. A.x(x = y -> ph) -> (x = y -> -. ph)))
7	1, 2, 6	19.21ad 1059	. . . 4 |- (-. A.x x = y -> (-. A.x(x = y -> ph) -> A.x(x = y -> -. ph)))
8		exanali 1043	. . . 4 |- (E.x(x = y /\ -. ph) <-> -. A.x(x = y -> ph))
9	7, 8	syl5ib 206	. . 3 |- (-. A.x x = y -> (E.x(x = y /\ -. ph) -> A.x(x = y -> -. ph)))
10		19.8a 1029	. . 3 |- ((x = y /\ -. ph) -> E.x(x = y /\ -. ph))
11	9, 10	syl5 21	. 2 |- (-. A.x x = y -> ((x = y /\ -. ph) -> A.x(x = y -> -. ph)))
12	11	exp3a 375	1 |- (-. A.x x = y -> (x = y -> (-. ph -> A.x(x = y -> -. ph))))

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223  A.wal 954   = wceq 956  E.wex 980

This theorem is referenced by:  ax11indi 1367  a12studyALT 1379

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-4 973  ax-5o 975  ax-6o 978

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981

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