Theorem ax11inda2 1761

Description: Induction step for constructing a substitution instance of ax-11o 1588 without using ax-11o 1588. Quantification case. When z and y are distinct, this theorem avoids the dummy variables needed by the more general ax11inda 1762.

Hypothesis

Ref	Expression
ax11inda2.1	|- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))

Assertion

Ref	Expression
ax11inda2	|- (-. A.x x = y -> (x = y -> (A.zph -> A.x(x = y -> A.zph))))

Distinct variable group:   y,z

Proof of Theorem ax11inda2

Step	Hyp	Ref	Expression
1		a16g 1653	. . . . 5 |- (A.y y = z -> ((x = y -> A.zph) -> A.x(x = y -> A.zph)))
2		ax-1 4	. . . . 5 |- (A.zph -> (x = y -> A.zph))
3	1, 2	syl5 20	. . . 4 |- (A.y y = z -> (A.zph -> A.x(x = y -> A.zph)))
4	3	a1d 15	. . 3 |- (A.y y = z -> (x = y -> (A.zph -> A.x(x = y -> A.zph))))
5	4	a1d 15	. 2 |- (A.y y = z -> (-. A.x x = y -> (x = y -> (A.zph -> A.x(x = y -> A.zph)))))
6		ax11inda2.1	. . 3 |- (-. A.x x = y -> (x = y -> (ph -> A.x(x = y -> ph))))
7	6	ax11indalem 1759	. 2 |- (-. A.y y = z -> (-. A.x x = y -> (x = y -> (A.zph -> A.x(x = y -> A.zph)))))
8	5, 7	pm2.61i 140	1 |- (-. A.x x = y -> (x = y -> (A.zph -> A.x(x = y -> A.zph))))

Syntax hints:  -. wn 2   -> wi 3  A.wal 1296   = wceq 1298

This theorem is referenced by:  ax11inda 1762

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-10 1308  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580

This theorem depends on definitions:  df-bi 164  df-an 242

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