BNF grammar associativity

I'm trying to understand how left and right associative grammars work and I need a little help. So I decided to come up an example and ask for some clarification. Basically, I want to create a grammar for two logical operations: and + implication. I want to make it so and is left associative and implication is right associative. This is what I got so far. Is this correct? I feel like it might be ambiguous. (I also kept in mind that and has higher precedence than implication)

<exp> := <and>
<and> := <impl> | <and> ^ <impl>
<impl> := <term> | <term> -> <impl>
<term> := (<exp>) | <bool>
<bool> := true | false

From my limited knowledge, it seems to me that you got the precedences inverted.

At the grammar level, a left associative operator has the following format:

exp = exp op other | other

...and a right associative operator would have the following format:

exp = other op exp | other

As you can see, it depends on your use of recursion: left associativity would use a left recursive rule while right associativity would use a right recursive one.

As for precedence, the later a rule is in the grammar, the higher its precedence. In the grammar bellow, where opL represents a left-associative operator and opR represents a right associative one, exp0 has lower precedence than exp1, which has lower precendence than other:

exp0 = exp0 opL exp1 | exp1
exp1 = other opR exp1 | other
other = ...

As an example, if opL is "+" and opR is "**" and other is a letter, see how the parse tree for a few expressions would be built:

• Left associativity:

  a + b + c -> (a + b) + c
  exp0 -+-> exp0 +-> exp0 --> exp1 --> other --> a
        |        |
        |        +-> opL --> "+"
        |        |
        |        \-> exp1 --> other --> b
        |
        +-> opL --> "+"
        |
        \-> exp1 --> c        
  

• Right Associativity:

    a ** b ** c -> a ** (b ** c)
    exp0 --> exp1 +-> other --> a
                  |
                  +-> opR --> "**"
                  |
                  \-> exp1 +-> other --> b
                           |
                           +-> opR --> "**"
                           |
                           \-> exp1 --> other --> c
  

• Precedence:

  a + b ** c -> a + (b ** c)
  exp0 +-> exp0 +-> exp1 --> other --> a
       |
       +-> opL --> "+"
       | 
       \-> exp1 +-> other --> b
                |
                +-> opR --> "**"
                |
                \-> exp1 --> other --> c