Metamath Proof Explorer


Theorem relexpdmg

Description: The domain of an exponentiation of a relation a subset of the relation's field. (Contributed by RP, 23-May-2020)

Ref Expression
Assertion relexpdmg ( ( 𝑁 ∈ ℕ0𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) )

Proof

Step Hyp Ref Expression
1 elnn0 ( 𝑁 ∈ ℕ0 ↔ ( 𝑁 ∈ ℕ ∨ 𝑁 = 0 ) )
2 relexpnndm ( ( 𝑁 ∈ ℕ ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ dom 𝑅 )
3 ssun1 dom 𝑅 ⊆ ( dom 𝑅 ∪ ran 𝑅 )
4 2 3 sstrdi ( ( 𝑁 ∈ ℕ ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) )
5 4 ex ( 𝑁 ∈ ℕ → ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) ) )
6 simpl ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → 𝑁 = 0 )
7 6 oveq2d ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → ( 𝑅𝑟 𝑁 ) = ( 𝑅𝑟 0 ) )
8 relexp0g ( 𝑅𝑉 → ( 𝑅𝑟 0 ) = ( I ↾ ( dom 𝑅 ∪ ran 𝑅 ) ) )
9 8 adantl ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → ( 𝑅𝑟 0 ) = ( I ↾ ( dom 𝑅 ∪ ran 𝑅 ) ) )
10 7 9 eqtrd ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → ( 𝑅𝑟 𝑁 ) = ( I ↾ ( dom 𝑅 ∪ ran 𝑅 ) ) )
11 10 dmeqd ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) = dom ( I ↾ ( dom 𝑅 ∪ ran 𝑅 ) ) )
12 dmresi dom ( I ↾ ( dom 𝑅 ∪ ran 𝑅 ) ) = ( dom 𝑅 ∪ ran 𝑅 )
13 11 12 eqtrdi ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) = ( dom 𝑅 ∪ ran 𝑅 ) )
14 eqimss ( dom ( 𝑅𝑟 𝑁 ) = ( dom 𝑅 ∪ ran 𝑅 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) )
15 13 14 syl ( ( 𝑁 = 0 ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) )
16 15 ex ( 𝑁 = 0 → ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) ) )
17 5 16 jaoi ( ( 𝑁 ∈ ℕ ∨ 𝑁 = 0 ) → ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) ) )
18 1 17 sylbi ( 𝑁 ∈ ℕ0 → ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) ) )
19 18 imp ( ( 𝑁 ∈ ℕ0𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ ( dom 𝑅 ∪ ran 𝑅 ) )