Theorem oprabidlem 5561

Description: Slight elaboration of exdistrfor 1722. A lemma for oprabid 5562. (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 1440	. . 3 ⊢ (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2		ax-10 1437	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3		dtru 4305	. . . . . 6 ⊢ ¬ ∀𝑦 𝑦 = 𝑧
4		pm2.53 674	. . . . . 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 1391	. . . . . 6 ⊢ (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
7	6	albii 1400	. . . . 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 709	. . 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 1722	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ 𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   ∨ wo 662  ∀wal 1283  Ⅎwnf 1390  ∃wex 1422

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 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-14 1446  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064  ax-sep 3898  ax-pow 3950  ax-setind 4282

This theorem depends on definitions:  df-bi 115  df-3an 922  df-tru 1288  df-fal 1291  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-ne 2247  df-ral 2354  df-v 2604  df-dif 2976  df-in 2980  df-ss 2987  df-pw 3386  df-sn 3406

This theorem is referenced by:  oprabid  5562