Metamath Proof Explorer


Theorem f1ssf1

Description: A subset of an injective function is injective. (Contributed by AV, 20-Nov-2020)

Ref Expression
Assertion f1ssf1 ( ( Fun 𝐹 ∧ Fun 𝐹𝐺𝐹 ) → Fun 𝐺 )

Proof

Step Hyp Ref Expression
1 funssres ( ( Fun 𝐹𝐺𝐹 ) → ( 𝐹 ↾ dom 𝐺 ) = 𝐺 )
2 funres11 ( Fun 𝐹 → Fun ( 𝐹 ↾ dom 𝐺 ) )
3 cnveq ( 𝐺 = ( 𝐹 ↾ dom 𝐺 ) → 𝐺 = ( 𝐹 ↾ dom 𝐺 ) )
4 3 funeqd ( 𝐺 = ( 𝐹 ↾ dom 𝐺 ) → ( Fun 𝐺 ↔ Fun ( 𝐹 ↾ dom 𝐺 ) ) )
5 2 4 syl5ibr ( 𝐺 = ( 𝐹 ↾ dom 𝐺 ) → ( Fun 𝐹 → Fun 𝐺 ) )
6 5 eqcoms ( ( 𝐹 ↾ dom 𝐺 ) = 𝐺 → ( Fun 𝐹 → Fun 𝐺 ) )
7 1 6 syl ( ( Fun 𝐹𝐺𝐹 ) → ( Fun 𝐹 → Fun 𝐺 ) )
8 7 ex ( Fun 𝐹 → ( 𝐺𝐹 → ( Fun 𝐹 → Fun 𝐺 ) ) )
9 8 com23 ( Fun 𝐹 → ( Fun 𝐹 → ( 𝐺𝐹 → Fun 𝐺 ) ) )
10 9 3imp ( ( Fun 𝐹 ∧ Fun 𝐹𝐺𝐹 ) → Fun 𝐺 )