Metamath Proof Explorer


Theorem func1st2nd

Description: Rewrite the functor predicate with separated parts. (Contributed by Zhi Wang, 19-Oct-2025)

Ref Expression
Hypothesis func1st2nd.1 ⊢ φ → F ∈ C Func D
Assertion func1st2nd ⊢ φ → 1 st ⁡ F C Func D 2 nd ⁡ F

Proof

Step Hyp Ref Expression
1 func1st2nd.1 ⊢ φ → F ∈ C Func D
2 relfunc ⊢ Rel ⁡ C Func D
3 1st2ndbr ⊢ Rel ⁡ C Func D ∧ F ∈ C Func D → 1 st ⁡ F C Func D 2 nd ⁡ F
4 2 1 3 sylancr ⊢ φ → 1 st ⁡ F C Func D 2 nd ⁡ F