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Pinkgo (토론 | 기여)님의 2025년 9월 23일 (화) 02:57 판 (→Ex. 3)

Ex. 1

Existence)
Let w be any string over an Σ.
By the definition, the domain of w is a finite initial segment of ℕ.
This means that ∃n∈ℕ s.t. w is defined at i↔i<n

Uniquness)
Suppose that ∃n,m s.t. both satisfy the condition for the string w.
That means, w is defined at position i if and only if (i<n)∧(i<m).
By definition, {i∈ℕ|i<n} and {i∈ℕ|i<m} are both domain of w, which means they are same set.
So the only way that {i∈ℕ|i<n} and {i∈ℕ|i<m} are same is their endpoints are same. Therefore, n=m.

Hence, there exists a unique natural number n with the desired property.

Ex. 2

Existence)
By definition, the domain of w is the finite initial segment of ℕ.
If the domain is ∅, then there is no position i∈ℕ where the w is defined.
In this situation, the length of the string w is 0.
Hence, there is a string w s.t. |w|=0.

Uniquness)
Suppose that there is two sets u,v s.t. |u|=0,|v|=0.
Then u,v are not defined at any index, so both are empty relation, subset of ℕ×Σ.
So u,v are both empty, which means u=v.

Hence, there is the a unique empty string w s.t. |w|=0, and we can define ϵ as the empty string w

Ex. 3

Let w be any string over an Σ.
Existence)
By the definition of w, w is defined at any position i s.t. (i∈ℕ)∧(i∈|w|).
Therefore, if i<|w|, then ∃x∈Σ.(i,x)∈w.

Uniquness)
By the definition of w, relation w⊆ℕ×Σ is partial function.
So there is only one output x∈Σ when i is given s.t. i<|w|∧i∈ℕ.

Hence, For all strings w and for all i∈ℕ, if i<|w|, then there exists a unique x∈Σ s.t. (i,x)∈w.

Ex. 4