Theorem axi12 1404

Description: Proof that ax-i12 1395 follows from ax-bnd 1396. So that we can track which theorems rely on ax-bnd 1396, proofs should reference ax-i12 1395 rather than this theorem. (Contributed by Jim Kingdon, 17-Aug-2018.) (New usage is discouraged). (Proof modification is discouraged.)

Assertion

Ref	Expression
axi12	|- (A.z z = x \/ (A.z z = y \/ A.z(x = y -> A.z x = y)))

Proof of Theorem axi12

Step	Hyp	Ref	Expression
1		ax-bnd 1396	. 2 |- (A.z z = x \/ (A.z z = y \/ A.xA.z(x = y -> A.z x = y)))
2		sp 1398	. . . 4 |- (A.xA.z(x = y -> A.z x = y) -> A.z(x = y -> A.z x = y))
3	2	orim2i 677	. . 3 |- ((A.z z = y \/ A.xA.z(x = y -> A.z x = y)) -> (A.z z = y \/ A.z(x = y -> A.z x = y)))
4	3	orim2i 677	. 2 |- ((A.z z = x \/ (A.z z = y \/ A.xA.z(x = y -> A.z x = y))) -> (A.z z = x \/ (A.z z = y \/ A.z(x = y -> A.z x = y))))
5	1, 4	ax-mp 7	1 |- (A.z z = x \/ (A.z z = y \/ A.z(x = y -> A.z x = y)))

Syntax hints:   -> wi 4   \/ wo 628  A.wal 1240   = wceq 1242

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-bnd 1396  ax-4 1397

This theorem depends on definitions:  df-bi 110

This theorem is referenced by: (None)