Metamath Proof Explorer


Theorem eldifsucnn

Description: Condition for membership in the difference of _om and a nonzero finite ordinal. (Contributed by Scott Fenton, 24-Oct-2024)

Ref Expression
Assertion eldifsucnn ⊢ A ∈ ω → B ∈ ω ∖ suc ⁡ A ↔ ∃ x ∈ ω ∖ A B = suc ⁡ x

Proof

Step Hyp Ref Expression
1 peano2 ⊢ A ∈ ω → suc ⁡ A ∈ ω
2 nnawordex ⊢ suc ⁡ A ∈ ω ∧ B ∈ ω → suc ⁡ A ⊆ B ↔ ∃ y ∈ ω suc ⁡ A + 𝑜 y = B
3 1 2 sylan ⊢ A ∈ ω ∧ B ∈ ω → suc ⁡ A ⊆ B ↔ ∃ y ∈ ω suc ⁡ A + 𝑜 y = B
4 nnacl ⊢ A ∈ ω ∧ y ∈ ω → A + 𝑜 y ∈ ω
5 nnaword1 ⊢ A ∈ ω ∧ y ∈ ω → A ⊆ A + 𝑜 y
6 nnasuc ⊢ y ∈ ω ∧ A ∈ ω → y + 𝑜 suc ⁡ A = suc ⁡ y + 𝑜 A
7 6 ancoms ⊢ A ∈ ω ∧ y ∈ ω → y + 𝑜 suc ⁡ A = suc ⁡ y + 𝑜 A
8 nnacom ⊢ suc ⁡ A ∈ ω ∧ y ∈ ω → suc ⁡ A + 𝑜 y = y + 𝑜 suc ⁡ A
9 1 8 sylan ⊢ A ∈ ω ∧ y ∈ ω → suc ⁡ A + 𝑜 y = y + 𝑜 suc ⁡ A
10 nnacom ⊢ A ∈ ω ∧ y ∈ ω → A + 𝑜 y = y + 𝑜 A
11 suceq ⊢ A + 𝑜 y = y + 𝑜 A → suc ⁡ A + 𝑜 y = suc ⁡ y + 𝑜 A
12 10 11 syl ⊢ A ∈ ω ∧ y ∈ ω → suc ⁡ A + 𝑜 y = suc ⁡ y + 𝑜 A
13 7 9 12 3eqtr4d ⊢ A ∈ ω ∧ y ∈ ω → suc ⁡ A + 𝑜 y = suc ⁡ A + 𝑜 y
14 sseq2 ⊢ x = A + 𝑜 y → A ⊆ x ↔ A ⊆ A + 𝑜 y
15 suceq ⊢ x = A + 𝑜 y → suc ⁡ x = suc ⁡ A + 𝑜 y
16 15 eqeq2d ⊢ x = A + 𝑜 y → suc ⁡ A + 𝑜 y = suc ⁡ x ↔ suc ⁡ A + 𝑜 y = suc ⁡ A + 𝑜 y
17 14 16 anbi12d ⊢ x = A + 𝑜 y → A ⊆ x ∧ suc ⁡ A + 𝑜 y = suc ⁡ x ↔ A ⊆ A + 𝑜 y ∧ suc ⁡ A + 𝑜 y = suc ⁡ A + 𝑜 y
18 17 rspcev ⊢ A + 𝑜 y ∈ ω ∧ A ⊆ A + 𝑜 y ∧ suc ⁡ A + 𝑜 y = suc ⁡ A + 𝑜 y → ∃ x ∈ ω A ⊆ x ∧ suc ⁡ A + 𝑜 y = suc ⁡ x
19 4 5 13 18 syl12anc ⊢ A ∈ ω ∧ y ∈ ω → ∃ x ∈ ω A ⊆ x ∧ suc ⁡ A + 𝑜 y = suc ⁡ x
20 eqeq1 ⊢ suc ⁡ A + 𝑜 y = B → suc ⁡ A + 𝑜 y = suc ⁡ x ↔ B = suc ⁡ x
21 20 anbi2d ⊢ suc ⁡ A + 𝑜 y = B → A ⊆ x ∧ suc ⁡ A + 𝑜 y = suc ⁡ x ↔ A ⊆ x ∧ B = suc ⁡ x
22 21 rexbidv ⊢ suc ⁡ A + 𝑜 y = B → ∃ x ∈ ω A ⊆ x ∧ suc ⁡ A + 𝑜 y = suc ⁡ x ↔ ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
23 19 22 syl5ibcom ⊢ A ∈ ω ∧ y ∈ ω → suc ⁡ A + 𝑜 y = B → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
24 23 rexlimdva ⊢ A ∈ ω → ∃ y ∈ ω suc ⁡ A + 𝑜 y = B → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
25 24 adantr ⊢ A ∈ ω ∧ B ∈ ω → ∃ y ∈ ω suc ⁡ A + 𝑜 y = B → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
26 3 25 sylbid ⊢ A ∈ ω ∧ B ∈ ω → suc ⁡ A ⊆ B → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
27 26 expimpd ⊢ A ∈ ω → B ∈ ω ∧ suc ⁡ A ⊆ B → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
28 peano2 ⊢ x ∈ ω → suc ⁡ x ∈ ω
29 28 ad2antlr ⊢ A ∈ ω ∧ x ∈ ω ∧ A ⊆ x → suc ⁡ x ∈ ω
30 nnord ⊢ A ∈ ω → Ord ⁡ A
31 nnord ⊢ x ∈ ω → Ord ⁡ x
32 ordsucsssuc ⊢ Ord ⁡ A ∧ Ord ⁡ x → A ⊆ x ↔ suc ⁡ A ⊆ suc ⁡ x
33 30 31 32 syl2an ⊢ A ∈ ω ∧ x ∈ ω → A ⊆ x ↔ suc ⁡ A ⊆ suc ⁡ x
34 33 biimpa ⊢ A ∈ ω ∧ x ∈ ω ∧ A ⊆ x → suc ⁡ A ⊆ suc ⁡ x
35 29 34 jca ⊢ A ∈ ω ∧ x ∈ ω ∧ A ⊆ x → suc ⁡ x ∈ ω ∧ suc ⁡ A ⊆ suc ⁡ x
36 eleq1 ⊢ B = suc ⁡ x → B ∈ ω ↔ suc ⁡ x ∈ ω
37 sseq2 ⊢ B = suc ⁡ x → suc ⁡ A ⊆ B ↔ suc ⁡ A ⊆ suc ⁡ x
38 36 37 anbi12d ⊢ B = suc ⁡ x → B ∈ ω ∧ suc ⁡ A ⊆ B ↔ suc ⁡ x ∈ ω ∧ suc ⁡ A ⊆ suc ⁡ x
39 35 38 syl5ibrcom ⊢ A ∈ ω ∧ x ∈ ω ∧ A ⊆ x → B = suc ⁡ x → B ∈ ω ∧ suc ⁡ A ⊆ B
40 39 expimpd ⊢ A ∈ ω ∧ x ∈ ω → A ⊆ x ∧ B = suc ⁡ x → B ∈ ω ∧ suc ⁡ A ⊆ B
41 40 rexlimdva ⊢ A ∈ ω → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x → B ∈ ω ∧ suc ⁡ A ⊆ B
42 27 41 impbid ⊢ A ∈ ω → B ∈ ω ∧ suc ⁡ A ⊆ B ↔ ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
43 eldif ⊢ B ∈ ω ∖ suc ⁡ A ↔ B ∈ ω ∧ ¬ B ∈ suc ⁡ A
44 nnord ⊢ suc ⁡ A ∈ ω → Ord ⁡ suc ⁡ A
45 1 44 syl ⊢ A ∈ ω → Ord ⁡ suc ⁡ A
46 nnord ⊢ B ∈ ω → Ord ⁡ B
47 ordtri1 ⊢ Ord ⁡ suc ⁡ A ∧ Ord ⁡ B → suc ⁡ A ⊆ B ↔ ¬ B ∈ suc ⁡ A
48 45 46 47 syl2an ⊢ A ∈ ω ∧ B ∈ ω → suc ⁡ A ⊆ B ↔ ¬ B ∈ suc ⁡ A
49 48 pm5.32da ⊢ A ∈ ω → B ∈ ω ∧ suc ⁡ A ⊆ B ↔ B ∈ ω ∧ ¬ B ∈ suc ⁡ A
50 43 49 bitr4id ⊢ A ∈ ω → B ∈ ω ∖ suc ⁡ A ↔ B ∈ ω ∧ suc ⁡ A ⊆ B
51 eldif ⊢ x ∈ ω ∖ A ↔ x ∈ ω ∧ ¬ x ∈ A
52 51 anbi1i ⊢ x ∈ ω ∖ A ∧ B = suc ⁡ x ↔ x ∈ ω ∧ ¬ x ∈ A ∧ B = suc ⁡ x
53 anass ⊢ x ∈ ω ∧ ¬ x ∈ A ∧ B = suc ⁡ x ↔ x ∈ ω ∧ ¬ x ∈ A ∧ B = suc ⁡ x
54 52 53 bitri ⊢ x ∈ ω ∖ A ∧ B = suc ⁡ x ↔ x ∈ ω ∧ ¬ x ∈ A ∧ B = suc ⁡ x
55 54 rexbii2 ⊢ ∃ x ∈ ω ∖ A B = suc ⁡ x ↔ ∃ x ∈ ω ¬ x ∈ A ∧ B = suc ⁡ x
56 ordtri1 ⊢ Ord ⁡ A ∧ Ord ⁡ x → A ⊆ x ↔ ¬ x ∈ A
57 30 31 56 syl2an ⊢ A ∈ ω ∧ x ∈ ω → A ⊆ x ↔ ¬ x ∈ A
58 57 anbi1d ⊢ A ∈ ω ∧ x ∈ ω → A ⊆ x ∧ B = suc ⁡ x ↔ ¬ x ∈ A ∧ B = suc ⁡ x
59 58 rexbidva ⊢ A ∈ ω → ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x ↔ ∃ x ∈ ω ¬ x ∈ A ∧ B = suc ⁡ x
60 55 59 bitr4id ⊢ A ∈ ω → ∃ x ∈ ω ∖ A B = suc ⁡ x ↔ ∃ x ∈ ω A ⊆ x ∧ B = suc ⁡ x
61 42 50 60 3bitr4d ⊢ A ∈ ω → B ∈ ω ∖ suc ⁡ A ↔ ∃ x ∈ ω ∖ A B = suc ⁡ x