Metamath Proof Explorer


Theorem fpr

Description: A function with a domain of two elements. (Contributed by Jeff Madsen, 20-Jun-2010) (Proof shortened by Andrew Salmon, 22-Oct-2011)

Ref Expression
Hypotheses fpr.1 ⊢ A ∈ V
fpr.2 ⊢ B ∈ V
fpr.3 ⊢ C ∈ V
fpr.4 ⊢ D ∈ V
Assertion fpr ⊢ A ≠ B → A C B D : A B ⟶ C D

Proof

Step Hyp Ref Expression
1 fpr.1 ⊢ A ∈ V
2 fpr.2 ⊢ B ∈ V
3 fpr.3 ⊢ C ∈ V
4 fpr.4 ⊢ D ∈ V
5 1 2 3 4 funpr ⊢ A ≠ B → Fun ⁡ A C B D
6 3 4 dmprop ⊢ dom ⁡ A C B D = A B
7 df-fn ⊢ A C B D Fn A B ↔ Fun ⁡ A C B D ∧ dom ⁡ A C B D = A B
8 5 6 7 sylanblrc ⊢ A ≠ B → A C B D Fn A B
9 df-pr ⊢ A C B D = A C ∪ B D
10 9 rneqi ⊢ ran ⁡ A C B D = ran ⁡ A C ∪ B D
11 rnun ⊢ ran ⁡ A C ∪ B D = ran ⁡ A C ∪ ran ⁡ B D
12 1 rnsnop ⊢ ran ⁡ A C = C
13 2 rnsnop ⊢ ran ⁡ B D = D
14 12 13 uneq12i ⊢ ran ⁡ A C ∪ ran ⁡ B D = C ∪ D
15 df-pr ⊢ C D = C ∪ D
16 14 15 eqtr4i ⊢ ran ⁡ A C ∪ ran ⁡ B D = C D
17 10 11 16 3eqtri ⊢ ran ⁡ A C B D = C D
18 17 eqimssi ⊢ ran ⁡ A C B D ⊆ C D
19 df-f ⊢ A C B D : A B ⟶ C D ↔ A C B D Fn A B ∧ ran ⁡ A C B D ⊆ C D
20 8 18 19 sylanblrc ⊢ A ≠ B → A C B D : A B ⟶ C D