Metamath Proof Explorer


Theorem rightval

Description: The value of the right options function. (Contributed by Scott Fenton, 9-Oct-2024)

Ref Expression
Assertion rightval ⊢ R ⁡ A = x ∈ Old ⁡ bday ⁡ A | A < s x

Proof

Step Hyp Ref Expression
1 2fveq3 ⊢ y = A → Old ⁡ bday ⁡ y = Old ⁡ bday ⁡ A
2 breq1 ⊢ y = A → y < s x ↔ A < s x
3 1 2 rabeqbidv ⊢ y = A → x ∈ Old ⁡ bday ⁡ y | y < s x = x ∈ Old ⁡ bday ⁡ A | A < s x
4 df-right ⊢ R = y ∈ No ⟼ x ∈ Old ⁡ bday ⁡ y | y < s x
5 fvex ⊢ Old ⁡ bday ⁡ A ∈ V
6 5 rabex ⊢ x ∈ Old ⁡ bday ⁡ A | A < s x ∈ V
7 3 4 6 fvmpt ⊢ A ∈ No → R ⁡ A = x ∈ Old ⁡ bday ⁡ A | A < s x
8 4 fvmptndm ⊢ ¬ A ∈ No → R ⁡ A = ∅
9 bdaydm ⊢ dom ⁡ bday = No
10 9 eleq2i ⊢ A ∈ dom ⁡ bday ↔ A ∈ No
11 ndmfv ⊢ ¬ A ∈ dom ⁡ bday → bday ⁡ A = ∅
12 10 11 sylnbir ⊢ ¬ A ∈ No → bday ⁡ A = ∅
13 12 fveq2d ⊢ ¬ A ∈ No → Old ⁡ bday ⁡ A = Old ⁡ ∅
14 old0 ⊢ Old ⁡ ∅ = ∅
15 13 14 eqtrdi ⊢ ¬ A ∈ No → Old ⁡ bday ⁡ A = ∅
16 15 rabeqdv ⊢ ¬ A ∈ No → x ∈ Old ⁡ bday ⁡ A | A < s x = x ∈ ∅ | A < s x
17 rab0 ⊢ x ∈ ∅ | A < s x = ∅
18 16 17 eqtrdi ⊢ ¬ A ∈ No → x ∈ Old ⁡ bday ⁡ A | A < s x = ∅
19 8 18 eqtr4d ⊢ ¬ A ∈ No → R ⁡ A = x ∈ Old ⁡ bday ⁡ A | A < s x
20 7 19 pm2.61i ⊢ R ⁡ A = x ∈ Old ⁡ bday ⁡ A | A < s x