-- Another example of a simple composition: compose p with its inverse
compinv (A : U) (a b : A) (p : Path A a b) : Path A a a =
 <i> comp (<_> A) (p @ i) [ (i = 0) -> <j> a, (i = 1) -> <j> p @ -j ]

-- Exercise (hard): is "compinv A a b p" Path equal to <j> a?
ex (A : U) (a b : A) (p : Path A a b) :
  Path (Path A a a) (compinv A a b p) (<i> a) =
    <i j> comp (<_> A) (p @ i \/ j) {- bottom -} [
      (i = 0) -> <k> p @ j /\ -k, -- back
      (i = 1) -> <k> p @      -k, -- front
      (j = 0) -> <k> p @ i /\ -k, -- left
      (j = 1) -> <k> p @      -k, -- right
    ]

-- Type checking failed: path endpoints don't match for <j> comp (<_> A) (p @ (i \/ j)) [ (i = 0) -> <k> p @ (j /\ -k), (i = 1) -> <k> p @ -k, (j = 0) -> <k> p @ (i /\ -k), (j = 1) -> <k> p @ -k ], got (<!0> a,<!0> a), but expected (<!0> comp (<!1> A) (p @ !0) [ (!0 = 0) -> <!1> a, (!0 = 1) -> <!1> p @ -!1 ],<!0> a)