Metamath Proof Explorer


Theorem rngoisoval

Description: Obsolete theorem, use rimval instead. The set of ring isomorphisms. (Contributed by Jeff Madsen, 16-Jun-2011) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypotheses rngisoval.1 ⊢ G = 1 st ⁡ R
rngisoval.2 ⊢ X = ran ⁡ G
rngisoval.3 ⊢ J = 1 st ⁡ S
rngisoval.4 ⊢ Y = ran ⁡ J
Assertion rngoisoval ⊢ R ∈ RingOps ∧ S ∈ RingOps → R RingOpsIso S = f ∈ R RingOpsHom S | f : X ⟶ 1-1 onto Y

Proof

Step Hyp Ref Expression
1 rngisoval.1 ⊢ G = 1 st ⁡ R
2 rngisoval.2 ⊢ X = ran ⁡ G
3 rngisoval.3 ⊢ J = 1 st ⁡ S
4 rngisoval.4 ⊢ Y = ran ⁡ J
5 oveq12 ⊢ r = R ∧ s = S → r RingOpsHom s = R RingOpsHom S
6 fveq2 ⊢ r = R → 1 st ⁡ r = 1 st ⁡ R
7 6 1 eqtr4di ⊢ r = R → 1 st ⁡ r = G
8 7 rneqd ⊢ r = R → ran ⁡ 1 st ⁡ r = ran ⁡ G
9 8 2 eqtr4di ⊢ r = R → ran ⁡ 1 st ⁡ r = X
10 9 f1oeq2d ⊢ r = R → f : ran ⁡ 1 st ⁡ r ⟶ 1-1 onto ran ⁡ 1 st ⁡ s ↔ f : X ⟶ 1-1 onto ran ⁡ 1 st ⁡ s
11 fveq2 ⊢ s = S → 1 st ⁡ s = 1 st ⁡ S
12 11 3 eqtr4di ⊢ s = S → 1 st ⁡ s = J
13 12 rneqd ⊢ s = S → ran ⁡ 1 st ⁡ s = ran ⁡ J
14 13 4 eqtr4di ⊢ s = S → ran ⁡ 1 st ⁡ s = Y
15 14 f1oeq3d ⊢ s = S → f : X ⟶ 1-1 onto ran ⁡ 1 st ⁡ s ↔ f : X ⟶ 1-1 onto Y
16 10 15 sylan9bb ⊢ r = R ∧ s = S → f : ran ⁡ 1 st ⁡ r ⟶ 1-1 onto ran ⁡ 1 st ⁡ s ↔ f : X ⟶ 1-1 onto Y
17 5 16 rabeqbidv ⊢ r = R ∧ s = S → f ∈ r RingOpsHom s | f : ran ⁡ 1 st ⁡ r ⟶ 1-1 onto ran ⁡ 1 st ⁡ s = f ∈ R RingOpsHom S | f : X ⟶ 1-1 onto Y
18 df-rngoiso ⊢ RingOpsIso = r ∈ RingOps , s ∈ RingOps ⟼ f ∈ r RingOpsHom s | f : ran ⁡ 1 st ⁡ r ⟶ 1-1 onto ran ⁡ 1 st ⁡ s
19 ovex ⊢ R RingOpsHom S ∈ V
20 19 rabex ⊢ f ∈ R RingOpsHom S | f : X ⟶ 1-1 onto Y ∈ V
21 17 18 20 ovmpoa ⊢ R ∈ RingOps ∧ S ∈ RingOps → R RingOpsIso S = f ∈ R RingOpsHom S | f : X ⟶ 1-1 onto Y