IfE and I are finite alphabets, define a string homomorphism to be a function : *...

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IfE and I are finite alphabets, define a string homomorphism to be a function : Σ* → I* that has the property that for any u, v € Σ*, φuυ) = φ(u) φ(υ). For example, the function : {0,..., 9, A,...,F}* → {0, 1}* that converts a hexadecimal number with n ≥ 0 digits into a binary number with 4n bits is a string homomorphism: (ε) = € E (0) = 0000 Φ(Α) = 1010 (CAB) 110010101011 = Intuitively, a string homomorphism does a "search and replace" where each symbol is replaced with a (possibly empty) string. Prove this more formally: that is, prove that if is a string homomorphism, then for any w = W₁ Wn (where n ≥ 0 and w; € Σ for 1 ≤ j≤n), we have (w) = (w₁) (wn). Use induction on n. (a) State and prove the base case (n = 0). (b) Assume that (*) is true for n = i and prove (*) for n = i + 1. ... (*) Activ Go to IfE and I are finite alphabets, define a string homomorphism to be a function : Σ* → I* that has the property that for any u, v € Σ*, φuυ) = φ(u) φ(υ). For example, the function : {0,..., 9, A,...,F}* → {0, 1}* that converts a hexadecimal number with n ≥ 0 digits into a binary number with 4n bits is a string homomorphism: (ε) = € E (0) = 0000 Φ(Α) = 1010 (CAB) 110010101011 = Intuitively, a string homomorphism does a "search and replace" where each symbol is replaced with a (possibly empty) string. Prove this more formally: that is, prove that if is a string homomorphism, then for any w = W₁ Wn (where n ≥ 0 and w; € Σ for 1 ≤ j≤n), we have (w) = (w₁) (wn). Use induction on n. (a) State and prove the base case (n = 0). (b) Assume that (*) is true for n = i and prove (*) for n = i + 1. ... (*) Activ Go to

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a State and prove the base case n 0 When n 0 we have w the empty string In this case we need to prov... View the full answer