Metamath Proof Explorer


Theorem isushgr

Description: The predicate "is an undirected simple hypergraph." (Contributed by AV, 19-Jan-2020) (Revised by AV, 9-Oct-2020)

Ref Expression
Hypotheses isuhgr.v V = Vtx G
isuhgr.e E = iEdg G
Assertion isushgr G U G USHGraph E : dom E 1-1 𝒫 V

Proof

Step Hyp Ref Expression
1 isuhgr.v V = Vtx G
2 isuhgr.e E = iEdg G
3 df-ushgr USHGraph = g | [˙ Vtx g / v]˙ [˙ iEdg g / e]˙ e : dom e 1-1 𝒫 v
4 3 eleq2i G USHGraph G g | [˙ Vtx g / v]˙ [˙ iEdg g / e]˙ e : dom e 1-1 𝒫 v
5 fveq2 h = G iEdg h = iEdg G
6 5 2 eqtr4di h = G iEdg h = E
7 5 dmeqd h = G dom iEdg h = dom iEdg G
8 2 eqcomi iEdg G = E
9 8 dmeqi dom iEdg G = dom E
10 7 9 eqtrdi h = G dom iEdg h = dom E
11 fveq2 h = G Vtx h = Vtx G
12 11 1 eqtr4di h = G Vtx h = V
13 12 pweqd h = G 𝒫 Vtx h = 𝒫 V
14 13 difeq1d h = G 𝒫 Vtx h = 𝒫 V
15 6 10 14 f1eq123d h = G iEdg h : dom iEdg h 1-1 𝒫 Vtx h E : dom E 1-1 𝒫 V
16 fvexd g = h Vtx g V
17 fveq2 g = h Vtx g = Vtx h
18 fvexd g = h v = Vtx h iEdg g V
19 fveq2 g = h iEdg g = iEdg h
20 19 adantr g = h v = Vtx h iEdg g = iEdg h
21 simpr g = h v = Vtx h e = iEdg h e = iEdg h
22 21 dmeqd g = h v = Vtx h e = iEdg h dom e = dom iEdg h
23 simpr g = h v = Vtx h v = Vtx h
24 23 pweqd g = h v = Vtx h 𝒫 v = 𝒫 Vtx h
25 24 difeq1d g = h v = Vtx h 𝒫 v = 𝒫 Vtx h
26 25 adantr g = h v = Vtx h e = iEdg h 𝒫 v = 𝒫 Vtx h
27 21 22 26 f1eq123d g = h v = Vtx h e = iEdg h e : dom e 1-1 𝒫 v iEdg h : dom iEdg h 1-1 𝒫 Vtx h
28 18 20 27 sbcied2 g = h v = Vtx h [˙ iEdg g / e]˙ e : dom e 1-1 𝒫 v iEdg h : dom iEdg h 1-1 𝒫 Vtx h
29 16 17 28 sbcied2 g = h [˙ Vtx g / v]˙ [˙ iEdg g / e]˙ e : dom e 1-1 𝒫 v iEdg h : dom iEdg h 1-1 𝒫 Vtx h
30 29 cbvabv g | [˙ Vtx g / v]˙ [˙ iEdg g / e]˙ e : dom e 1-1 𝒫 v = h | iEdg h : dom iEdg h 1-1 𝒫 Vtx h
31 15 30 elab2g G U G g | [˙ Vtx g / v]˙ [˙ iEdg g / e]˙ e : dom e 1-1 𝒫 v E : dom E 1-1 𝒫 V
32 4 31 syl5bb G U G USHGraph E : dom E 1-1 𝒫 V