Metamath Proof Explorer


Theorem noetasuplem2

Description: Lemma for noeta . The restriction of Z to dom S is S . (Contributed by Scott Fenton, 9-Aug-2024)

Ref Expression
Hypotheses noetasuplem.1 S = if x A y A ¬ x < s y ι x A | y A ¬ x < s y dom ι x A | y A ¬ x < s y 2 𝑜 g y | u A y dom u v A ¬ v < s u u suc y = v suc y ι x | u A g dom u v A ¬ v < s u u suc g = v suc g u g = x
noetasuplem.2 Z = S suc bday B dom S × 1 𝑜
Assertion noetasuplem2 A No A V B V Z dom S = S

Proof

Step Hyp Ref Expression
1 noetasuplem.1 S = if x A y A ¬ x < s y ι x A | y A ¬ x < s y dom ι x A | y A ¬ x < s y 2 𝑜 g y | u A y dom u v A ¬ v < s u u suc y = v suc y ι x | u A g dom u v A ¬ v < s u u suc g = v suc g u g = x
2 noetasuplem.2 Z = S suc bday B dom S × 1 𝑜
3 2 reseq1i Z dom S = S suc bday B dom S × 1 𝑜 dom S
4 resundir S suc bday B dom S × 1 𝑜 dom S = S dom S suc bday B dom S × 1 𝑜 dom S
5 dmres dom suc bday B dom S × 1 𝑜 dom S = dom S dom suc bday B dom S × 1 𝑜
6 1oex 1 𝑜 V
7 6 snnz 1 𝑜
8 dmxp 1 𝑜 dom suc bday B dom S × 1 𝑜 = suc bday B dom S
9 7 8 ax-mp dom suc bday B dom S × 1 𝑜 = suc bday B dom S
10 9 ineq2i dom S dom suc bday B dom S × 1 𝑜 = dom S suc bday B dom S
11 disjdif dom S suc bday B dom S =
12 5 10 11 3eqtri dom suc bday B dom S × 1 𝑜 dom S =
13 relres Rel suc bday B dom S × 1 𝑜 dom S
14 reldm0 Rel suc bday B dom S × 1 𝑜 dom S suc bday B dom S × 1 𝑜 dom S = dom suc bday B dom S × 1 𝑜 dom S =
15 13 14 ax-mp suc bday B dom S × 1 𝑜 dom S = dom suc bday B dom S × 1 𝑜 dom S =
16 12 15 mpbir suc bday B dom S × 1 𝑜 dom S =
17 16 uneq2i S dom S suc bday B dom S × 1 𝑜 dom S = S dom S
18 3 4 17 3eqtri Z dom S = S dom S
19 1 nosupno A No A V S No
20 19 3adant3 A No A V B V S No
21 nofun S No Fun S
22 20 21 syl A No A V B V Fun S
23 funrel Fun S Rel S
24 resdm Rel S S dom S = S
25 22 23 24 3syl A No A V B V S dom S = S
26 25 uneq1d A No A V B V S dom S = S
27 un0 S = S
28 26 27 eqtrdi A No A V B V S dom S = S
29 18 28 syl5eq A No A V B V Z dom S = S