Metamath Proof Explorer


Theorem noseponlem

Description: Lemma for nosepon . Consider a case of proper subset domain. (Contributed by Scott Fenton, 21-Sep-2020)

Ref Expression
Assertion noseponlem ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ¬ ∀ 𝑥 ∈ On ( 𝐴𝑥 ) = ( 𝐵𝑥 ) )

Proof

Step Hyp Ref Expression
1 nodmon ( 𝐴 No → dom 𝐴 ∈ On )
2 1 3ad2ant1 ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → dom 𝐴 ∈ On )
3 nodmord ( 𝐴 No → Ord dom 𝐴 )
4 ordirr ( Ord dom 𝐴 → ¬ dom 𝐴 ∈ dom 𝐴 )
5 3 4 syl ( 𝐴 No → ¬ dom 𝐴 ∈ dom 𝐴 )
6 5 3ad2ant1 ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ¬ dom 𝐴 ∈ dom 𝐴 )
7 ndmfv ( ¬ dom 𝐴 ∈ dom 𝐴 → ( 𝐴 ‘ dom 𝐴 ) = ∅ )
8 6 7 syl ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ( 𝐴 ‘ dom 𝐴 ) = ∅ )
9 nosgnn0 ¬ ∅ ∈ { 1o , 2o }
10 elno3 ( 𝐵 No ↔ ( 𝐵 : dom 𝐵 ⟶ { 1o , 2o } ∧ dom 𝐵 ∈ On ) )
11 10 simplbi ( 𝐵 No 𝐵 : dom 𝐵 ⟶ { 1o , 2o } )
12 11 3ad2ant2 ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → 𝐵 : dom 𝐵 ⟶ { 1o , 2o } )
13 simp3 ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → dom 𝐴 ∈ dom 𝐵 )
14 12 13 ffvelrnd ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ( 𝐵 ‘ dom 𝐴 ) ∈ { 1o , 2o } )
15 eleq1 ( ( 𝐵 ‘ dom 𝐴 ) = ∅ → ( ( 𝐵 ‘ dom 𝐴 ) ∈ { 1o , 2o } ↔ ∅ ∈ { 1o , 2o } ) )
16 14 15 syl5ibcom ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ( ( 𝐵 ‘ dom 𝐴 ) = ∅ → ∅ ∈ { 1o , 2o } ) )
17 9 16 mtoi ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ¬ ( 𝐵 ‘ dom 𝐴 ) = ∅ )
18 17 neqned ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ( 𝐵 ‘ dom 𝐴 ) ≠ ∅ )
19 18 necomd ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ∅ ≠ ( 𝐵 ‘ dom 𝐴 ) )
20 8 19 eqnetrd ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ( 𝐴 ‘ dom 𝐴 ) ≠ ( 𝐵 ‘ dom 𝐴 ) )
21 fveq2 ( 𝑥 = dom 𝐴 → ( 𝐴𝑥 ) = ( 𝐴 ‘ dom 𝐴 ) )
22 fveq2 ( 𝑥 = dom 𝐴 → ( 𝐵𝑥 ) = ( 𝐵 ‘ dom 𝐴 ) )
23 21 22 neeq12d ( 𝑥 = dom 𝐴 → ( ( 𝐴𝑥 ) ≠ ( 𝐵𝑥 ) ↔ ( 𝐴 ‘ dom 𝐴 ) ≠ ( 𝐵 ‘ dom 𝐴 ) ) )
24 23 rspcev ( ( dom 𝐴 ∈ On ∧ ( 𝐴 ‘ dom 𝐴 ) ≠ ( 𝐵 ‘ dom 𝐴 ) ) → ∃ 𝑥 ∈ On ( 𝐴𝑥 ) ≠ ( 𝐵𝑥 ) )
25 2 20 24 syl2anc ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ∃ 𝑥 ∈ On ( 𝐴𝑥 ) ≠ ( 𝐵𝑥 ) )
26 df-ne ( ( 𝐴𝑥 ) ≠ ( 𝐵𝑥 ) ↔ ¬ ( 𝐴𝑥 ) = ( 𝐵𝑥 ) )
27 26 rexbii ( ∃ 𝑥 ∈ On ( 𝐴𝑥 ) ≠ ( 𝐵𝑥 ) ↔ ∃ 𝑥 ∈ On ¬ ( 𝐴𝑥 ) = ( 𝐵𝑥 ) )
28 rexnal ( ∃ 𝑥 ∈ On ¬ ( 𝐴𝑥 ) = ( 𝐵𝑥 ) ↔ ¬ ∀ 𝑥 ∈ On ( 𝐴𝑥 ) = ( 𝐵𝑥 ) )
29 27 28 bitri ( ∃ 𝑥 ∈ On ( 𝐴𝑥 ) ≠ ( 𝐵𝑥 ) ↔ ¬ ∀ 𝑥 ∈ On ( 𝐴𝑥 ) = ( 𝐵𝑥 ) )
30 25 29 sylib ( ( 𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵 ) → ¬ ∀ 𝑥 ∈ On ( 𝐴𝑥 ) = ( 𝐵𝑥 ) )