Annotated parse tree for $(3+4)\times(5+6)$

Syntax-directed definition (synthesized val):

• $L \to E$, $\ L.val = E.val$
• $E \to T$, $E.val = T.val$
• $E \to E_1+T$, $E.val = E_1.val + T.val$
• $T \to F$, $T.val = F.val$
• $T \to T_1*F$, $T.val = T_1.val \times F.val$
• $F \to (E)$, $F.val = E.val$
• $F \to \text{digit}$, $F.val = \text{digit.lexval}$

Parse tree with every node's val annotated (evaluated bottom-up):

              L ─ E.val = 77                    (L → E, L.val = E.val)
                  |
                  E.val = 77                    (E → T,   E.val = T.val)
                  |
                  T.val = 77                    (T → T1 * F,  T.val = 7 * 11)
                 /         \
        T1.val = 7          F.val = 11
             |                    |
             F.val = 7           F → ( E )  .val = 11
             F → ( E ) .val=7         |
                |                    E → E1 + T  .val = 5 + 6 = 11
                E.val = 7            /        \
              E1.val = 3          T.val = 3   ?  ... (details below)

To avoid a confusing drawing, here is the node-by-node annotated tree (the expected exam answer):

L → E            : val = 77
E → T            : val = 77
T → T1 * F       : val = T1.val * F.val = 7 * 11 = 77

Left subtree  (3+4):
  T1 → F                          : val = 7
  F → (E)                         : val = 7      [parenthesized]
     E → E1 + T
        E1 → T → F → digit(3)     : val = 3
        T  → F → digit(4)         : val = 4
        E.val = 3 + 4 = 7
  ⇒ T1 (left factor) val = 7

Right subtree  (5+6):
  F → ( E )                       : val = 11     [parenthesized]
     E → E1 + T
        E1 → T → F → digit(5)     : val = 5
        T → F → digit(6)          : val = 6
        E.val = 5 + 6 = 11
  ⇒ F (right factor) val = 11

Root:
  T.val = 7 * 11 = 77,  E.val = 77,  L.val = 77

Each digit is a leaf via $F\to\text{digit}$, with $F.val = \text{digit.lexval}$ (3, 4, 5, 6).

$$\boxed{E.val = 7 \times 11 = 77}$$