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$$H(m)=E(K, IV \oplus m) = C_1$$
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$$H(m)=E(K, IV \oplus m) = C_1$$
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$$H(m')=E(K, IV \oplus m') = C_2$$
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$$H(m')=E(K, IV \oplus m') = C_2$$
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And if there the hashing function was not collision resistant that would imply
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And if there the hashing function was not collision resistant that would imply
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$$C_1=C_2\implies D(C_1)=D(C_2) \impies m=m'$$
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$$C_1=C_2\implies D(C_1)=D(C_2) \implies m=m'$$
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and since $m !=m'$ the hash function is collision resistant, for messages with 1 block.
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and since $m !=m'$ the hash function is collision resistant, for messages with 1 block.
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