Question

student be allowed to add any explanations to his/her answer after it has been submitted. Problem 1 (20 points): Show that for any two regular expressions r and s the following is true: $(r^s^)^* = (r \cup s)^$ For example, taking r = 1011(0\cup 1) and s = (1\cup \varepsilon)11(0\cup 1), the above equality implies that $( (10^11(0\cup 1))^* ((1\cup \varepsilon)11^(0\cup 1))^ )^* = ( (10^11(0\cup 1)) \cup ((1\cup \varepsilon)11^(0\cup 1)))^$ r s r s Note that proof by example and proof by vague-o-logy (i.e., vague descriptions) are not valid proof methods. Hint: Show that $(r^s^)^ \subseteq (r+s)^$ and $(r^s^)^ \supseteq (r+s)^$. In other words, show that: • If a string is in $(r^s^)^$ then the string is also in $(r+s)^$. • If a string is in $(r+s)^$ then the string is also in $(r^s^)^*.$

          student be allowed to add any explanations to his/her answer after it has been submitted.
Problem 1 (20 points): Show that for any two regular expressions r and s the following is true:
$(r^*s^*)^* = (r \cup s)^*$
For example, taking r = 10*11(0\cup 1) and s = (1\cup \varepsilon)11*(0\cup 1), the above equality implies
that
$( (10^*11(0\cup 1))^* ((1\cup \varepsilon)11^*(0\cup 1))^* )^* = ( (10^*11(0\cup 1)) \cup ((1\cup \varepsilon)11^*(0\cup 1)))^*$
r
s
r
s
Note that proof by example and proof by vague-o-logy (i.e., vague descriptions) are
not valid proof methods.
Hint: Show that $(r^*s^*)^* \subseteq (r+s)^*$ and $(r^*s^*)^* \supseteq (r+s)^*$. In other words, show that:
• If a string is in $(r^*s^*)^*$ then the string is also in $(r+s)^*$.
• If a string is in $(r+s)^*$ then the string is also in $(r^*s^*)^*.$

student be allowed to add any explanations to his/her answer after it has been submitted.
Problem 1 (20 points): Show that for any two regular expressions r and s the following is true:
(r^*s^*)^* = (r ∪ s)^*
For example, taking r = 10*11(0∪1) and s = (1∪ε)11*(0∪1), the above equality implies
that
( (10^*11(0∪ 1))^* ((1∪ε)11^*(0∪ 1))^* )^* = ( (10^*11(0∪ 1)) ∪ ((1∪ε)11^*(0∪ 1)))^*
r
s
r
s
Note that proof by example and proof by vague-o-logy (i.e., vague descriptions) are
not valid proof methods.
Hint: Show that (r^*s^*)^* ⊆ (r+s)^* and (r^*s^*)^* ⊇ (r+s)^*. In other words, show that:
• If a string is in (r^*s^*)^* then the string is also in (r+s)^*.
• If a string is in (r+s)^* then the string is also in (r^*s^*)^*.

Added by Ruben J.

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