Step |
Hyp |
Ref |
Expression |
1 |
|
fndmdif |
⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → dom ( 𝐹 ∖ 𝐺 ) = { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } ) |
2 |
1
|
eqeq1d |
⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( dom ( 𝐹 ∖ 𝐺 ) = ∅ ↔ { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ) ) |
3 |
|
rabeq0 |
⊢ ( { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ↔ ∀ 𝑥 ∈ 𝐴 ¬ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) ) |
4 |
|
nne |
⊢ ( ¬ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) ↔ ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) |
5 |
4
|
ralbii |
⊢ ( ∀ 𝑥 ∈ 𝐴 ¬ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) |
6 |
3 5
|
bitri |
⊢ ( { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) |
7 |
|
eqfnfv |
⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) ) |
8 |
6 7
|
bitr4id |
⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ↔ 𝐹 = 𝐺 ) ) |
9 |
2 8
|
bitrd |
⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( dom ( 𝐹 ∖ 𝐺 ) = ∅ ↔ 𝐹 = 𝐺 ) ) |