Metamath Proof Explorer


Theorem extdg1b

Description: The degree of the extension E /FldExt F is 1 iff E and F are the same structure. (Contributed by Thierry Arnoux, 6-Aug-2023)

Ref Expression
Assertion extdg1b ⊢ E /FldExt F → E .:. F = 1 ↔ E = F

Proof

Step Hyp Ref Expression
1 extdg1id ⊢ E /FldExt F ∧ E .:. F = 1 → E = F
2 oveq1 ⊢ E = F → E .:. F = F .:. F
3 2 adantl ⊢ E /FldExt F ∧ E = F → E .:. F = F .:. F
4 fldextfld2 ⊢ E /FldExt F → F ∈ Field
5 4 adantr ⊢ E /FldExt F ∧ E = F → F ∈ Field
6 extdgid ⊢ F ∈ Field → F .:. F = 1
7 5 6 syl ⊢ E /FldExt F ∧ E = F → F .:. F = 1
8 3 7 eqtrd ⊢ E /FldExt F ∧ E = F → E .:. F = 1
9 1 8 impbida ⊢ E /FldExt F → E .:. F = 1 ↔ E = F