Theorem axi12 1404

Description: Proof that ax-i12 1395 follows from ax-bndl 1396. So that we can track which theorems rely on ax-bndl 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	⊢ (∀z z = x ∨ (∀z z = y ∨ ∀z(x = y → ∀z x = y)))

Proof of Theorem axi12

Step	Hyp	Ref	Expression
1		ax-bndl 1396	. 2 ⊢ (∀z z = x ∨ (∀z z = y ∨ ∀x∀z(x = y → ∀z x = y)))
2		sp 1398	. . . 4 ⊢ (∀x∀z(x = y → ∀z x = y) → ∀z(x = y → ∀z x = y))
3	2	orim2i 677	. . 3 ⊢ ((∀z z = y ∨ ∀x∀z(x = y → ∀z x = y)) → (∀z z = y ∨ ∀z(x = y → ∀z x = y)))
4	3	orim2i 677	. 2 ⊢ ((∀z z = x ∨ (∀z z = y ∨ ∀x∀z(x = y → ∀z x = y))) → (∀z z = x ∨ (∀z z = y ∨ ∀z(x = y → ∀z x = y))))
5	1, 4	ax-mp 7	1 ⊢ (∀z z = x ∨ (∀z z = y ∨ ∀z(x = y → ∀z x = y)))

Syntax hints:   → wi 4   ∨ wo 628  ∀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-bndl 1396  ax-4 1397

This theorem depends on definitions:  df-bi 110

This theorem is referenced by: (None)