Metamath Proof Explorer


Theorem fnsuppeq0

Description: The support of a function is empty iff it is identically zero. (Contributed by Stefan O'Rear, 22-Mar-2015) (Revised by AV, 28-May-2019)

Ref Expression
Assertion fnsuppeq0 ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( ( 𝐹 supp 𝑍 ) = ∅ ↔ 𝐹 = ( 𝐴 × { 𝑍 } ) ) )

Proof

Step Hyp Ref Expression
1 ss0b ( ( 𝐹 supp 𝑍 ) ⊆ ∅ ↔ ( 𝐹 supp 𝑍 ) = ∅ )
2 un0 ( 𝐴 ∪ ∅ ) = 𝐴
3 uncom ( 𝐴 ∪ ∅ ) = ( ∅ ∪ 𝐴 )
4 2 3 eqtr3i 𝐴 = ( ∅ ∪ 𝐴 )
5 4 fneq2i ( 𝐹 Fn 𝐴𝐹 Fn ( ∅ ∪ 𝐴 ) )
6 5 biimpi ( 𝐹 Fn 𝐴𝐹 Fn ( ∅ ∪ 𝐴 ) )
7 6 3ad2ant1 ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → 𝐹 Fn ( ∅ ∪ 𝐴 ) )
8 fnex ( ( 𝐹 Fn 𝐴𝐴𝑊 ) → 𝐹 ∈ V )
9 8 3adant3 ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → 𝐹 ∈ V )
10 simp3 ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → 𝑍𝑉 )
11 0in ( ∅ ∩ 𝐴 ) = ∅
12 11 a1i ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( ∅ ∩ 𝐴 ) = ∅ )
13 fnsuppres ( ( 𝐹 Fn ( ∅ ∪ 𝐴 ) ∧ ( 𝐹 ∈ V ∧ 𝑍𝑉 ) ∧ ( ∅ ∩ 𝐴 ) = ∅ ) → ( ( 𝐹 supp 𝑍 ) ⊆ ∅ ↔ ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) ) )
14 7 9 10 12 13 syl121anc ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( ( 𝐹 supp 𝑍 ) ⊆ ∅ ↔ ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) ) )
15 1 14 syl5bbr ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( ( 𝐹 supp 𝑍 ) = ∅ ↔ ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) ) )
16 fnresdm ( 𝐹 Fn 𝐴 → ( 𝐹𝐴 ) = 𝐹 )
17 16 3ad2ant1 ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( 𝐹𝐴 ) = 𝐹 )
18 17 eqeq1d ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) ↔ 𝐹 = ( 𝐴 × { 𝑍 } ) ) )
19 15 18 bitrd ( ( 𝐹 Fn 𝐴𝐴𝑊𝑍𝑉 ) → ( ( 𝐹 supp 𝑍 ) = ∅ ↔ 𝐹 = ( 𝐴 × { 𝑍 } ) ) )