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Theorem axi12 1447
Description: Proof that ax-i12 1438 follows from ax-bndl 1439. So that we can track which theorems rely on ax-bndl 1439, proofs should reference ax-i12 1438 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
StepHypRef Expression
1 ax-bndl 1439 . 2 (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
2 sp 1441 . . . 4 (∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦) → ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))
32orim2i 710 . . 3 ((∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)) → (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
43orim2i 710 . 2 ((∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑥𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))) → (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦))))
51, 4ax-mp 7 1 (∀𝑧 𝑧 = 𝑥 ∨ (∀𝑧 𝑧 = 𝑦 ∨ ∀𝑧(𝑥 = 𝑦 → ∀𝑧 𝑥 = 𝑦)))
Colors of variables: wff set class
Syntax hints:  wi 4  wo 661  wal 1282   = wceq 1284
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-bndl 1439  ax-4 1440
This theorem depends on definitions:  df-bi 115
This theorem is referenced by: (None)
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