scala - Restrict method of a trait with constraint on abstract type member using implicits? -

i in situation below:

import scalaz.leibniz._  trait exp[t, c] {     def &&(that: exp[t, c])(implicit evt: t === boolean) = logicaland(this, that)    def &&(that: exp[t, c])(implicit evt: t === int) = bitwiseand(this, that)  }  case class logicaland[c](e1: exp[boolean, c], e2: exp[boolean, c]) extends exp[boolean, c]  case class logicalor[c](e1: exp[boolean, c], e2: exp[boolean, c]) extends exp[boolean, c]  ... case class bitwiseand[c](e1: exp[int, c], e2: exp[int, c]) extends exp[int, c] case class bitwiseor[c](e1: exp[int, c], e2: exp[int, c]) extends exp[int, c] ... 

the trait exp[t,c] base trait , ast dsl, overload built-in scala operators in trait allow infix notation on dsl, constrain of these methods bound on type t @ trait level same operation here '&&' has different semantics depending on type t.

it seems leibniz subsitution not/cannot work here (maybe because defined functors f[_] single argument):

[error] /home/remi/projects/dsl/src/main/scala/exp.scala:80: type mismatch; [error]  found   : exp.exp[t,c] [error]  required: exp.exp[boolean,?] [error]   = logicaland(this, that) [error]         ^ 

does approach constraining trait's t parameter make sense @ ? there way make leibniz work in case "hiding" second parameter c :

type expf[t] = exp[t, _] 

if makes sense? thanks,

exactly that; define such type , use leibniz.lift to, well, lift leibniz:

def &&(that: exp[t, c])(implicit evt: t === boolean) = {   type expf[a] = exp[a, c]   val ev2: (exp[t, c] === exp[boolean, c]) =     leibniz.lift[⊥, ⊥, ⊤, ⊤, expf, t, boolean](evt)   logicaland(this, that) }