Theorem oprabidlem 5884

Description: Slight elaboration of exdistrfor 1793. A lemma for oprabid 5885. (Contributed by Jim Kingdon, 15-Jan-2019.)

Assertion

Ref	Expression
oprabidlem	⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ 𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧

Allowed substitution hints:   𝜓(𝑥,𝑦,𝑧)

Proof of Theorem oprabidlem

Step	Hyp	Ref	Expression
1		ax-bndl 1502	. . 3 ⊢ (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2		ax-10 1498	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3		dtru 4544	. . . . . 6 ⊢ ¬ ∀𝑦 𝑦 = 𝑧
4		pm2.53 717	. . . . . 6 ⊢ ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → (¬ ∀𝑦 𝑦 = 𝑧 → ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
5	3, 4	mpi 15	. . . . 5 ⊢ ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
6		df-nf 1454	. . . . . 6 ⊢ (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
7	6	albii 1463	. . . . 5 ⊢ (∀𝑥Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
8	5, 7	sylibr 133	. . . 4 ⊢ ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥Ⅎ𝑦 𝑥 = 𝑧)
9	2, 8	orim12i 754	. . 3 ⊢ ((∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))) → (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥Ⅎ𝑦 𝑥 = 𝑧))
10	1, 9	ax-mp 5	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥Ⅎ𝑦 𝑥 = 𝑧)
11	10	exdistrfor 1793	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ 𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ∨ wo 703  ∀wal 1346  Ⅎwnf 1453  ∃wex 1485

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-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-14 2144  ax-ext 2152  ax-sep 4107  ax-pow 4160  ax-setind 4521

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-fal 1354  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ne 2341  df-ral 2453  df-v 2732  df-dif 3123  df-in 3127  df-ss 3134  df-pw 3568  df-sn 3589

This theorem is referenced by:  oprabid  5885