For correct semantics, we express &&,|| using ?: operator:

x && y === x ? y : false

x || y === x ? true : y

Resulting translation:

[[ x && y ]] =
        [[ x ]]
        ifeq nFalse
        [[ y ]]
        goto nAfter
nFalse: iconst_0
nAfter: 

Applying the rule above, we obtain

[[ (x && y) && z ]] =
         [[ x ]]
         ifeq nFalse1
         [[ y ]]
         goto nAfter1
nFalse1: iconst_0
nAfter1: ifeq nFalse2
         [[ z ]]
         goto nAfter2
nFalse2: iconst_0
nAfter2: 

The above is correct

• example below is a little more compact

class Test {
    boolean and(boolean p, boolean q, boolean r) {
	return p && q && r;
    }
 
    boolean or(boolean p, boolean q, boolean r) {
	return p || q || r;
    }
}

Translation:

boolean and(boolean, boolean, boolean);
  Code:
   0:	iload_1
   1:	ifeq	16
   4:	iload_2
   5:	ifeq	16
   8:	iload_3
   9:	ifeq	16
   12:	iconst_1
   13:	goto	17
   16:	iconst_0
   17:	ireturn

boolean or(boolean, boolean, boolean);
  Code:
   0:	iload_1
   1:	ifne	12
   4:	iload_2
   5:	ifne	12
   8:	iload_3
   9:	ifeq	16
   12:	iconst_1
   13:	goto	17
   16:	iconst_0
   17:	ireturn