Theorem axi12 1502

Description: Proof that ax-i12 1495 follows from ax-bndl 1497. So that we can track which theorems rely on ax-bndl 1497, proofs should reference ax12or 1496 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 1497	. 2 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
2		sp 1499	. . . 4 ⊢ (∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
3	2	orim2i 751	. . 3 ⊢ ((∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
4	3	orim2i 751	. 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 5	1 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))

Syntax hints:   → wi 4   ∨ wo 698  ∀wal 1341   = wceq 1343

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-bndl 1497  ax-4 1498

This theorem depends on definitions:  df-bi 116

This theorem is referenced by: (None)