Theorem nfso 5500

Description: Bound-variable hypothesis builder for total orders. (Contributed by Stefan O'Rear, 20-Jan-2015.)

Hypotheses

Ref	Expression
nfpo.r	⊢ Ⅎ𝑥𝑅
nfpo.a	⊢ Ⅎ𝑥𝐴

Assertion

Ref	Expression
nfso	⊢ Ⅎ𝑥 𝑅 Or 𝐴

Proof of Theorem nfso

Dummy variables 𝑎 𝑏 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-so 5495	. 2 ⊢ (𝑅 Or 𝐴 ↔ (𝑅 Po 𝐴 ∧ ∀𝑎 ∈ 𝐴 ∀𝑏 ∈ 𝐴 (𝑎𝑅𝑏 ∨ 𝑎 = 𝑏 ∨ 𝑏𝑅𝑎)))
2		nfpo.r	. . . 4 ⊢ Ⅎ𝑥𝑅
3		nfpo.a	. . . 4 ⊢ Ⅎ𝑥𝐴
4	2, 3	nfpo 5499	. . 3 ⊢ Ⅎ𝑥 𝑅 Po 𝐴
5		nfcv 2906	. . . . . . 7 ⊢ Ⅎ𝑥𝑎
6		nfcv 2906	. . . . . . 7 ⊢ Ⅎ𝑥𝑏
7	5, 2, 6	nfbr 5117	. . . . . 6 ⊢ Ⅎ𝑥 𝑎𝑅𝑏
8		nfv 1918	. . . . . 6 ⊢ Ⅎ𝑥 𝑎 = 𝑏
9	6, 2, 5	nfbr 5117	. . . . . 6 ⊢ Ⅎ𝑥 𝑏𝑅𝑎
10	7, 8, 9	nf3or 1909	. . . . 5 ⊢ Ⅎ𝑥(𝑎𝑅𝑏 ∨ 𝑎 = 𝑏 ∨ 𝑏𝑅𝑎)
11	3, 10	nfralw 3149	. . . 4 ⊢ Ⅎ𝑥∀𝑏 ∈ 𝐴 (𝑎𝑅𝑏 ∨ 𝑎 = 𝑏 ∨ 𝑏𝑅𝑎)
12	3, 11	nfralw 3149	. . 3 ⊢ Ⅎ𝑥∀𝑎 ∈ 𝐴 ∀𝑏 ∈ 𝐴 (𝑎𝑅𝑏 ∨ 𝑎 = 𝑏 ∨ 𝑏𝑅𝑎)
13	4, 12	nfan 1903	. 2 ⊢ Ⅎ𝑥(𝑅 Po 𝐴 ∧ ∀𝑎 ∈ 𝐴 ∀𝑏 ∈ 𝐴 (𝑎𝑅𝑏 ∨ 𝑎 = 𝑏 ∨ 𝑏𝑅𝑎))
14	1, 13	nfxfr 1856	1 ⊢ Ⅎ𝑥 𝑅 Or 𝐴

Syntax hints:   ∧ wa 395   ∨ w3o 1084  Ⅎwnf 1787  Ⅎwnfc 2886  ∀wral 3063   class class class wbr 5070   Po wpo 5492   Or wor 5493

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3or 1086  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ral 3068  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-br 5071  df-po 5494  df-so 5495

This theorem is referenced by:  nfwe  5556