Metamath Proof Explorer


Theorem dicelval3

Description: Member of the partial isomorphism C. (Contributed by NM, 26-Feb-2014)

Ref Expression
Hypotheses dicval.l ⊢ ≤ ˙ = ≤ K
dicval.a ⊢ A = Atoms ⁡ K
dicval.h ⊢ H = LHyp ⁡ K
dicval.p ⊢ P = oc ⁡ K ⁡ W
dicval.t ⊢ T = LTrn ⁡ K ⁡ W
dicval.e ⊢ E = TEndo ⁡ K ⁡ W
dicval.i ⊢ I = DIsoC ⁡ K ⁡ W
dicval2.g ⊢ G = ι g ∈ T | g ⁡ P = Q
Assertion dicelval3 ⊢ K ∈ V ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W → Y ∈ I ⁡ Q ↔ ∃ s ∈ E Y = s ⁡ G s

Proof

Step Hyp Ref Expression
1 dicval.l ⊢ ≤ ˙ = ≤ K
2 dicval.a ⊢ A = Atoms ⁡ K
3 dicval.h ⊢ H = LHyp ⁡ K
4 dicval.p ⊢ P = oc ⁡ K ⁡ W
5 dicval.t ⊢ T = LTrn ⁡ K ⁡ W
6 dicval.e ⊢ E = TEndo ⁡ K ⁡ W
7 dicval.i ⊢ I = DIsoC ⁡ K ⁡ W
8 dicval2.g ⊢ G = ι g ∈ T | g ⁡ P = Q
9 1 2 3 4 5 6 7 8 dicval2 ⊢ K ∈ V ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W → I ⁡ Q = f s | f = s ⁡ G ∧ s ∈ E
10 9 eleq2d ⊢ K ∈ V ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W → Y ∈ I ⁡ Q ↔ Y ∈ f s | f = s ⁡ G ∧ s ∈ E
11 excom ⊢ ∃ f ∃ s Y = f s ∧ f = s ⁡ G ∧ s ∈ E ↔ ∃ s ∃ f Y = f s ∧ f = s ⁡ G ∧ s ∈ E
12 an12 ⊢ Y = f s ∧ f = s ⁡ G ∧ s ∈ E ↔ f = s ⁡ G ∧ Y = f s ∧ s ∈ E
13 12 exbii ⊢ ∃ f Y = f s ∧ f = s ⁡ G ∧ s ∈ E ↔ ∃ f f = s ⁡ G ∧ Y = f s ∧ s ∈ E
14 fvex ⊢ s ⁡ G ∈ V
15 opeq1 ⊢ f = s ⁡ G → f s = s ⁡ G s
16 15 eqeq2d ⊢ f = s ⁡ G → Y = f s ↔ Y = s ⁡ G s
17 16 anbi1d ⊢ f = s ⁡ G → Y = f s ∧ s ∈ E ↔ Y = s ⁡ G s ∧ s ∈ E
18 14 17 ceqsexv ⊢ ∃ f f = s ⁡ G ∧ Y = f s ∧ s ∈ E ↔ Y = s ⁡ G s ∧ s ∈ E
19 ancom ⊢ Y = s ⁡ G s ∧ s ∈ E ↔ s ∈ E ∧ Y = s ⁡ G s
20 13 18 19 3bitri ⊢ ∃ f Y = f s ∧ f = s ⁡ G ∧ s ∈ E ↔ s ∈ E ∧ Y = s ⁡ G s
21 20 exbii ⊢ ∃ s ∃ f Y = f s ∧ f = s ⁡ G ∧ s ∈ E ↔ ∃ s s ∈ E ∧ Y = s ⁡ G s
22 11 21 bitri ⊢ ∃ f ∃ s Y = f s ∧ f = s ⁡ G ∧ s ∈ E ↔ ∃ s s ∈ E ∧ Y = s ⁡ G s
23 elopab ⊢ Y ∈ f s | f = s ⁡ G ∧ s ∈ E ↔ ∃ f ∃ s Y = f s ∧ f = s ⁡ G ∧ s ∈ E
24 df-rex ⊢ ∃ s ∈ E Y = s ⁡ G s ↔ ∃ s s ∈ E ∧ Y = s ⁡ G s
25 22 23 24 3bitr4i ⊢ Y ∈ f s | f = s ⁡ G ∧ s ∈ E ↔ ∃ s ∈ E Y = s ⁡ G s
26 10 25 bitrdi ⊢ K ∈ V ∧ W ∈ H ∧ Q ∈ A ∧ ¬ Q ≤ ˙ W → Y ∈ I ⁡ Q ↔ ∃ s ∈ E Y = s ⁡ G s