EXAMPLE 3.2 - Using the grammar in Example 32, show a parse tree and a leftmost derivation for the following statement: A=Aast (B+(Cast A)) A Grammar for Simple Assignment Statements cassign>-dd>-cexpn <ib=Al B |C capp-cid>+expp | ( <expp) I dd

To show a parse tree and a leftmost derivation for the statement $A=A\ast (B+(C\ast A))$, we can use the grammar provided in Example 32. <br /><br />The grammar for simple assignment statements is as follows:<br />```<br /><assign> ::= <id> = <expn><br /><expn> ::= <id> | <app><br /><app> ::= <id> <op> <expn><br /><op> ::= +<br />```<br /><br />Here, `<assign>` represents an assignment statement, `<id>` represents an identifier, `<expn>` represents an expression, `<app>` represents an application of two expressions, and `<op>` represents an operator.<br /><br />To construct a parse tree, we start with the given statement and apply the grammar rules to break it down into its components. The parse tree for the statement $A=A\ast (B+(C\ast A))$ would look like this:<br /><br />```<br />A<br />├── A = A * (B + (C * A))<br />│ ├── A<br />│ ├── *<br />│ ├── B + (C * A)<br />│ │ ├── B<br />│ │ ├── +<br />│ │ ├── C * A<br />│ │ │ ├── C<br />│ │ │ ├── *<br />│ │ │ ├── A<br />```<br /><br />To construct a leftmost derivation, we start with the given statement and apply the grammar rules to generate a sequence of symbols that represents the statement. The leftmost derivation for the statement $A=A\ast (B+(C\ast A))$ would look like this:<br /><br />```<br />A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A => A (B + (C * A))<br /> => A * + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (BC * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br /> => A = A * (B + (C * A))<br />