Metamath Proof Explorer


Theorem ofsubeq0

Description: Function analogue of subeq0 . (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Assertion ofsubeq0 A V F : A G : A F f G = A × 0 F = G

Proof

Step Hyp Ref Expression
1 simp2 A V F : A G : A F : A
2 1 ffnd A V F : A G : A F Fn A
3 simp3 A V F : A G : A G : A
4 3 ffnd A V F : A G : A G Fn A
5 simp1 A V F : A G : A A V
6 inidm A A = A
7 eqidd A V F : A G : A x A F x = F x
8 eqidd A V F : A G : A x A G x = G x
9 2 4 5 5 6 7 8 ofval A V F : A G : A x A F f G x = F x G x
10 c0ex 0 V
11 10 fvconst2 x A A × 0 x = 0
12 11 adantl A V F : A G : A x A A × 0 x = 0
13 9 12 eqeq12d A V F : A G : A x A F f G x = A × 0 x F x G x = 0
14 1 ffvelrnda A V F : A G : A x A F x
15 3 ffvelrnda A V F : A G : A x A G x
16 14 15 subeq0ad A V F : A G : A x A F x G x = 0 F x = G x
17 13 16 bitrd A V F : A G : A x A F f G x = A × 0 x F x = G x
18 17 ralbidva A V F : A G : A x A F f G x = A × 0 x x A F x = G x
19 2 4 5 5 6 offn A V F : A G : A F f G Fn A
20 10 fconst A × 0 : A 0
21 ffn A × 0 : A 0 A × 0 Fn A
22 20 21 ax-mp A × 0 Fn A
23 eqfnfv F f G Fn A A × 0 Fn A F f G = A × 0 x A F f G x = A × 0 x
24 19 22 23 sylancl A V F : A G : A F f G = A × 0 x A F f G x = A × 0 x
25 eqfnfv F Fn A G Fn A F = G x A F x = G x
26 2 4 25 syl2anc A V F : A G : A F = G x A F x = G x
27 18 24 26 3bitr4d A V F : A G : A F f G = A × 0 F = G