Metamath Proof Explorer


Theorem fvunsn

Description: Remove an ordered pair not participating in a function value. (Contributed by NM, 1-Oct-2013) (Revised by Mario Carneiro, 28-May-2014)

Ref Expression
Assertion fvunsn ⊢ B ≠ D → A ∪ B C ⁡ D = A ⁡ D

Proof

Step Hyp Ref Expression
1 resundir ⊢ A ∪ B C ↾ D = A ↾ D ∪ B C ↾ D
2 nelsn ⊢ B ≠ D → ¬ B ∈ D
3 ressnop0 ⊢ ¬ B ∈ D → B C ↾ D = ∅
4 2 3 syl ⊢ B ≠ D → B C ↾ D = ∅
5 4 uneq2d ⊢ B ≠ D → A ↾ D ∪ B C ↾ D = A ↾ D ∪ ∅
6 un0 ⊢ A ↾ D ∪ ∅ = A ↾ D
7 5 6 eqtrdi ⊢ B ≠ D → A ↾ D ∪ B C ↾ D = A ↾ D
8 1 7 eqtrid ⊢ B ≠ D → A ∪ B C ↾ D = A ↾ D
9 8 fveq1d ⊢ B ≠ D → A ∪ B C ↾ D ⁡ D = A ↾ D ⁡ D
10 fvressn ⊢ D ∈ V → A ∪ B C ↾ D ⁡ D = A ∪ B C ⁡ D
11 fvprc ⊢ ¬ D ∈ V → A ∪ B C ↾ D ⁡ D = ∅
12 fvprc ⊢ ¬ D ∈ V → A ∪ B C ⁡ D = ∅
13 11 12 eqtr4d ⊢ ¬ D ∈ V → A ∪ B C ↾ D ⁡ D = A ∪ B C ⁡ D
14 10 13 pm2.61i ⊢ A ∪ B C ↾ D ⁡ D = A ∪ B C ⁡ D
15 fvressn ⊢ D ∈ V → A ↾ D ⁡ D = A ⁡ D
16 fvprc ⊢ ¬ D ∈ V → A ↾ D ⁡ D = ∅
17 fvprc ⊢ ¬ D ∈ V → A ⁡ D = ∅
18 16 17 eqtr4d ⊢ ¬ D ∈ V → A ↾ D ⁡ D = A ⁡ D
19 15 18 pm2.61i ⊢ A ↾ D ⁡ D = A ⁡ D
20 9 14 19 3eqtr3g ⊢ B ≠ D → A ∪ B C ⁡ D = A ⁡ D