Theorem axi12 1407

Description: Proof that ax-i12 1398 follows from ax-bndl 1399. So that we can track which theorems rely on ax-bndl 1399, proofs should reference ax-i12 1398 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 1399	. 2 ⊢ (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
2		sp 1401	. . . 4 ⊢ (∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
3	2	orim2i 678	. . 3 ⊢ ((∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
4	3	orim2i 678	. 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 629  ∀wal 1241   = wceq 1243

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 630  ax-bndl 1399  ax-4 1400

This theorem depends on definitions:  df-bi 110

This theorem is referenced by: (None)