Theorem oprabidlem 5988

Description: Slight elaboration of exdistrfor 1824. A lemma for oprabid 5989. (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 1533	. . 3 ⊢ (∀𝑦 𝑦 = 𝑥 ∨ (∀𝑦 𝑦 = 𝑧 ∨ ∀𝑥∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧)))
2		ax-10 1529	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑥 → ∀𝑥 𝑥 = 𝑦)
3		dtru 4616	. . . . . 6 ⊢ ¬ ∀𝑦 𝑦 = 𝑧
4		pm2.53 724	. . . . . 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 1485	. . . . . 6 ⊢ (Ⅎ𝑦 𝑥 = 𝑧 ↔ ∀𝑦(𝑥 = 𝑧 → ∀𝑦 𝑥 = 𝑧))
7	6	albii 1494	. . . . 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 761	. . 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 1824	1 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ 𝜓) → ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ∨ wo 710  ∀wal 1371  Ⅎwnf 1484  ∃wex 1516

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 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-14 2180  ax-ext 2188  ax-sep 4170  ax-pow 4226  ax-setind 4593

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-fal 1379  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-ne 2378  df-ral 2490  df-v 2775  df-dif 3172  df-in 3176  df-ss 3183  df-pw 3623  df-sn 3644

This theorem is referenced by:  oprabid  5989