Let be a finite alphabet. Two languages L1, L2 are many-one equivalent, denoted L1 m L2, if L1 m L2 and L2 m L1. L1 is a proper subset of L2, denoted L1 L2, if L1 L2 and L1 6= L2. Prove or disprove the following conjectures.
2 Conjecture. L1 RE and L2 RE and L1 and L2 = L1 m L2.
Solution
Given statement: L1 RE and L2 RE and L1 and L2 = L1 m L2
As in the question, we have clear explaination regarding many-one equivalent. be a finite alphabet. Two languages L1, L2 are many-one equivalent, denoted L1 m L2, if L1 m L2 and L2 m L1. L1 is a proper subset of L2, denoted L1 L2, if L1 L2 and L1 = L2.
So , As  L1 RE and L2 RE these are both subset of all finite alphabets. As both languages are proper subsets of finite alphabets. So they are many-one equivalent. So it holds the conjecture
L1 and L2 = L1 m L2
.
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Let be a finite alphabet- Two languages L1- L2 are many-one equiva.docx
1. Let be a finite alphabet. Two languages L1, L2 are many-one equivalent, denoted L1 m L2, if L1
m L2 and L2 m L1. L1 is a proper subset of L2, denoted L1 L2, if L1 L2 and L1 6= L2. Prove or
disprove the following conjectures.
2 Conjecture. L1 RE and L2 RE and L1 and L2 = L1 m L2.
Solution
Given statement: L1 RE and L2 RE and L1 and L2 = L1 m L2
As in the question, we have clear explaination regarding many-one equivalent. be a finite
alphabet. Two languages L1, L2 are many-one equivalent, denoted L1 m L2, if L1 m L2 and L2
m L1. L1 is a proper subset of L2, denoted L1 L2, if L1 L2 and L1 = L2.
So , As  L1 RE and L2 RE these are both subset of all finite alphabets. As both languages are
proper subsets of finite alphabets. So they are many-one equivalent. So it holds the conjecture
L1 and L2 = L1 m L2
