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