Theorem oprabidlem 5680

Description: Slight elaboration of exdistrfor 1728. A lemma for oprabid 5681. (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 1444	. . 3 ⊢ (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2		ax-10 1441	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3		dtru 4376	. . . . . 6 ⊢ ¬ ∀𝑦 𝑦 = 𝑧
4		pm2.53 676	. . . . . 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 1395	. . . . . 6 ⊢ (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
7	6	albii 1404	. . . . 5 ⊢ (∀𝑥Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
8	5, 7	sylibr 132	. . . 4 ⊢ ((∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)) → ∀𝑥Ⅎ𝑦 𝑥 = 𝑧)
9	2, 8	orim12i 711	. . 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 7	. 2 ⊢ (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥Ⅎ𝑦 𝑥 = 𝑧)
11	10	exdistrfor 1728	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ 𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   ∨ wo 664  ∀wal 1287  Ⅎwnf 1394  ∃wex 1426

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-in1 579  ax-in2 580  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-14 1450  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070  ax-sep 3957  ax-pow 4009  ax-setind 4353

This theorem depends on definitions:  df-bi 115  df-3an 926  df-tru 1292  df-fal 1295  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-ne 2256  df-ral 2364  df-v 2621  df-dif 3001  df-in 3005  df-ss 3012  df-pw 3431  df-sn 3452

This theorem is referenced by:  oprabid  5681