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:

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

The next rules are strictly ordered:

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

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