Theorem oprabidlem 6059

Description: Slight elaboration of exdistrfor 1848. A lemma for oprabid 6060. (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 1558	. . 3 ⊢ (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2		ax-10 1554	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3		dtru 4664	. . . . . 6 ⊢ ¬ ∀𝑦 𝑦 = 𝑧
4		pm2.53 730	. . . . . 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 1510	. . . . . 6 ⊢ (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
7	6	albii 1519	. . . . 5 ⊢ (∀𝑥Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
8	5, 7	sylibr 134	. . . 4 ⊢ ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥Ⅎ𝑦 𝑥 = 𝑧)
9	2, 8	orim12i 767	. . 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 1848	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ 𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ∨ wo 716  ∀wal 1396  Ⅎwnf 1509  ∃wex 1541

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-14 2205  ax-ext 2213  ax-sep 4212  ax-pow 4270  ax-setind 4641

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ne 2404  df-ral 2516  df-v 2805  df-dif 3203  df-in 3207  df-ss 3214  df-pw 3658  df-sn 3679

This theorem is referenced by:  oprabid  6060