Metamath Proof Explorer


Theorem relexpnndm

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 relexpnndm ( ( 𝑁 ∈ ℕ ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ dom 𝑅 )

Proof

Step Hyp Ref Expression
1 oveq2 ( 𝑛 = 1 → ( 𝑅𝑟 𝑛 ) = ( 𝑅𝑟 1 ) )
2 1 dmeqd ( 𝑛 = 1 → dom ( 𝑅𝑟 𝑛 ) = dom ( 𝑅𝑟 1 ) )
3 2 sseq1d ( 𝑛 = 1 → ( dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ↔ dom ( 𝑅𝑟 1 ) ⊆ dom 𝑅 ) )
4 3 imbi2d ( 𝑛 = 1 → ( ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ) ↔ ( 𝑅𝑉 → dom ( 𝑅𝑟 1 ) ⊆ dom 𝑅 ) ) )
5 oveq2 ( 𝑛 = 𝑚 → ( 𝑅𝑟 𝑛 ) = ( 𝑅𝑟 𝑚 ) )
6 5 dmeqd ( 𝑛 = 𝑚 → dom ( 𝑅𝑟 𝑛 ) = dom ( 𝑅𝑟 𝑚 ) )
7 6 sseq1d ( 𝑛 = 𝑚 → ( dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ↔ dom ( 𝑅𝑟 𝑚 ) ⊆ dom 𝑅 ) )
8 7 imbi2d ( 𝑛 = 𝑚 → ( ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ) ↔ ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑚 ) ⊆ dom 𝑅 ) ) )
9 oveq2 ( 𝑛 = ( 𝑚 + 1 ) → ( 𝑅𝑟 𝑛 ) = ( 𝑅𝑟 ( 𝑚 + 1 ) ) )
10 9 dmeqd ( 𝑛 = ( 𝑚 + 1 ) → dom ( 𝑅𝑟 𝑛 ) = dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) )
11 10 sseq1d ( 𝑛 = ( 𝑚 + 1 ) → ( dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ↔ dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) ⊆ dom 𝑅 ) )
12 11 imbi2d ( 𝑛 = ( 𝑚 + 1 ) → ( ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ) ↔ ( 𝑅𝑉 → dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) ⊆ dom 𝑅 ) ) )
13 oveq2 ( 𝑛 = 𝑁 → ( 𝑅𝑟 𝑛 ) = ( 𝑅𝑟 𝑁 ) )
14 13 dmeqd ( 𝑛 = 𝑁 → dom ( 𝑅𝑟 𝑛 ) = dom ( 𝑅𝑟 𝑁 ) )
15 14 sseq1d ( 𝑛 = 𝑁 → ( dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ↔ dom ( 𝑅𝑟 𝑁 ) ⊆ dom 𝑅 ) )
16 15 imbi2d ( 𝑛 = 𝑁 → ( ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑛 ) ⊆ dom 𝑅 ) ↔ ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑁 ) ⊆ dom 𝑅 ) ) )
17 relexp1g ( 𝑅𝑉 → ( 𝑅𝑟 1 ) = 𝑅 )
18 17 dmeqd ( 𝑅𝑉 → dom ( 𝑅𝑟 1 ) = dom 𝑅 )
19 eqimss ( dom ( 𝑅𝑟 1 ) = dom 𝑅 → dom ( 𝑅𝑟 1 ) ⊆ dom 𝑅 )
20 18 19 syl ( 𝑅𝑉 → dom ( 𝑅𝑟 1 ) ⊆ dom 𝑅 )
21 relexpsucnnr ( ( 𝑅𝑉𝑚 ∈ ℕ ) → ( 𝑅𝑟 ( 𝑚 + 1 ) ) = ( ( 𝑅𝑟 𝑚 ) ∘ 𝑅 ) )
22 21 ancoms ( ( 𝑚 ∈ ℕ ∧ 𝑅𝑉 ) → ( 𝑅𝑟 ( 𝑚 + 1 ) ) = ( ( 𝑅𝑟 𝑚 ) ∘ 𝑅 ) )
23 22 dmeqd ( ( 𝑚 ∈ ℕ ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) = dom ( ( 𝑅𝑟 𝑚 ) ∘ 𝑅 ) )
24 dmcoss dom ( ( 𝑅𝑟 𝑚 ) ∘ 𝑅 ) ⊆ dom 𝑅
25 23 24 eqsstrdi ( ( 𝑚 ∈ ℕ ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) ⊆ dom 𝑅 )
26 25 a1d ( ( 𝑚 ∈ ℕ ∧ 𝑅𝑉 ) → ( dom ( 𝑅𝑟 𝑚 ) ⊆ dom 𝑅 → dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) ⊆ dom 𝑅 ) )
27 26 ex ( 𝑚 ∈ ℕ → ( 𝑅𝑉 → ( dom ( 𝑅𝑟 𝑚 ) ⊆ dom 𝑅 → dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) ⊆ dom 𝑅 ) ) )
28 27 a2d ( 𝑚 ∈ ℕ → ( ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑚 ) ⊆ dom 𝑅 ) → ( 𝑅𝑉 → dom ( 𝑅𝑟 ( 𝑚 + 1 ) ) ⊆ dom 𝑅 ) ) )
29 4 8 12 16 20 28 nnind ( 𝑁 ∈ ℕ → ( 𝑅𝑉 → dom ( 𝑅𝑟 𝑁 ) ⊆ dom 𝑅 ) )
30 29 imp ( ( 𝑁 ∈ ℕ ∧ 𝑅𝑉 ) → dom ( 𝑅𝑟 𝑁 ) ⊆ dom 𝑅 )