YES

We show the termination of the TRS R:

  |:|(x,x) -> e()
  |:|(x,e()) -> x
  i(|:|(x,y)) -> |:|(y,x)
  |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
  |:|(e(),x) -> i(x)
  i(i(x)) -> x
  i(e()) -> e()
  |:|(x,|:|(y,i(x))) -> i(y)
  |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
  |:|(i(x),|:|(y,x)) -> i(y)
  |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))
p3: |:|#(|:|(x,y),z) -> |:|#(z,i(y))
p4: |:|#(|:|(x,y),z) -> i#(y)
p5: |:|#(e(),x) -> i#(x)
p6: |:|#(x,|:|(y,i(x))) -> i#(y)
p7: |:|#(x,|:|(y,|:|(i(x),z))) -> |:|#(i(z),y)
p8: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p9: |:|#(i(x),|:|(y,x)) -> i#(y)
p10: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p11: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: i#(|:|(x,y)) -> |:|#(y,x)
p2: |:|#(i(x),|:|(y,|:|(x,z))) -> i#(z)
p3: |:|#(i(x),|:|(y,|:|(x,z))) -> |:|#(i(z),y)
p4: |:|#(i(x),|:|(y,x)) -> i#(y)
p5: |:|#(x,|:|(y,|:|(i(x),z))) -> i#(z)
p6: |:|#(x,|:|(y,|:|(i(x),z))) -> |:|#(i(z),y)
p7: |:|#(x,|:|(y,i(x))) -> i#(y)
p8: |:|#(e(),x) -> i#(x)
p9: |:|#(|:|(x,y),z) -> i#(y)
p10: |:|#(|:|(x,y),z) -> |:|#(z,i(y))
p11: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      i#_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (0,0,0,1)
      |:|_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,0,0)) x1 + x2 + (2,2,1,1)
      |:|#_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2
      i_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + (0,4,3,3)
      e_A() = (1,1,4,4)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p9, p10

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: |:|#(e(),x) -> i#(x)
p2: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: |:|#(|:|(x,y),z) -> |:|#(x,|:|(z,i(y)))

and R consists of:

r1: |:|(x,x) -> e()
r2: |:|(x,e()) -> x
r3: i(|:|(x,y)) -> |:|(y,x)
r4: |:|(|:|(x,y),z) -> |:|(x,|:|(z,i(y)))
r5: |:|(e(),x) -> i(x)
r6: i(i(x)) -> x
r7: i(e()) -> e()
r8: |:|(x,|:|(y,i(x))) -> i(y)
r9: |:|(x,|:|(y,|:|(i(x),z))) -> |:|(i(z),y)
r10: |:|(i(x),|:|(y,x)) -> i(y)
r11: |:|(i(x),|:|(y,|:|(x,z))) -> |:|(i(z),y)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      |:|#_A(x1,x2) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,1,0,0)) x1
      |:|_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,1,0,0)) x2 + (2,2,1,4)
      i_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,1,0,0)) x1 + (0,1,7,1)
      e_A() = (1,5,2,5)

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.