Metamath Proof Explorer


Theorem f1opr

Description: Condition for an operation to be one-to-one. (Contributed by Jeff Madsen, 17-Jun-2010)

Ref Expression
Assertion f1opr ⊢ F : A × B ⟶ 1-1 C ↔ F : A × B ⟶ C ∧ ∀ r ∈ A ∀ s ∈ B ∀ t ∈ A ∀ u ∈ B r F s = t F u → r = t ∧ s = u

Proof

Step Hyp Ref Expression
1 dff13 ⊢ F : A × B ⟶ 1-1 C ↔ F : A × B ⟶ C ∧ ∀ v ∈ A × B ∀ w ∈ A × B F ⁡ v = F ⁡ w → v = w
2 fveq2 ⊢ v = r s → F ⁡ v = F ⁡ r s
3 df-ov ⊢ r F s = F ⁡ r s
4 2 3 eqtr4di ⊢ v = r s → F ⁡ v = r F s
5 4 eqeq1d ⊢ v = r s → F ⁡ v = F ⁡ w ↔ r F s = F ⁡ w
6 eqeq1 ⊢ v = r s → v = w ↔ r s = w
7 5 6 imbi12d ⊢ v = r s → F ⁡ v = F ⁡ w → v = w ↔ r F s = F ⁡ w → r s = w
8 7 ralbidv ⊢ v = r s → ∀ w ∈ A × B F ⁡ v = F ⁡ w → v = w ↔ ∀ w ∈ A × B r F s = F ⁡ w → r s = w
9 8 ralxp ⊢ ∀ v ∈ A × B ∀ w ∈ A × B F ⁡ v = F ⁡ w → v = w ↔ ∀ r ∈ A ∀ s ∈ B ∀ w ∈ A × B r F s = F ⁡ w → r s = w
10 fveq2 ⊢ w = t u → F ⁡ w = F ⁡ t u
11 df-ov ⊢ t F u = F ⁡ t u
12 10 11 eqtr4di ⊢ w = t u → F ⁡ w = t F u
13 12 eqeq2d ⊢ w = t u → r F s = F ⁡ w ↔ r F s = t F u
14 eqeq2 ⊢ w = t u → r s = w ↔ r s = t u
15 vex ⊢ r ∈ V
16 vex ⊢ s ∈ V
17 15 16 opth ⊢ r s = t u ↔ r = t ∧ s = u
18 14 17 bitrdi ⊢ w = t u → r s = w ↔ r = t ∧ s = u
19 13 18 imbi12d ⊢ w = t u → r F s = F ⁡ w → r s = w ↔ r F s = t F u → r = t ∧ s = u
20 19 ralxp ⊢ ∀ w ∈ A × B r F s = F ⁡ w → r s = w ↔ ∀ t ∈ A ∀ u ∈ B r F s = t F u → r = t ∧ s = u
21 20 2ralbii ⊢ ∀ r ∈ A ∀ s ∈ B ∀ w ∈ A × B r F s = F ⁡ w → r s = w ↔ ∀ r ∈ A ∀ s ∈ B ∀ t ∈ A ∀ u ∈ B r F s = t F u → r = t ∧ s = u
22 9 21 bitri ⊢ ∀ v ∈ A × B ∀ w ∈ A × B F ⁡ v = F ⁡ w → v = w ↔ ∀ r ∈ A ∀ s ∈ B ∀ t ∈ A ∀ u ∈ B r F s = t F u → r = t ∧ s = u
23 22 anbi2i ⊢ F : A × B ⟶ C ∧ ∀ v ∈ A × B ∀ w ∈ A × B F ⁡ v = F ⁡ w → v = w ↔ F : A × B ⟶ C ∧ ∀ r ∈ A ∀ s ∈ B ∀ t ∈ A ∀ u ∈ B r F s = t F u → r = t ∧ s = u
24 1 23 bitri ⊢ F : A × B ⟶ 1-1 C ↔ F : A × B ⟶ C ∧ ∀ r ∈ A ∀ s ∈ B ∀ t ∈ A ∀ u ∈ B r F s = t F u → r = t ∧ s = u