YES
We show the termination of the TRS R:

  .(|1|(),x) -> x
  .(x,|1|()) -> x
  .(i(x),x) -> |1|()
  .(x,i(x)) -> |1|()
  i(|1|()) -> |1|()
  i(i(x)) -> x
  .(i(y),.(y,z)) -> z
  .(y,.(i(y),z)) -> z
  .(.(x,y),z) -> .(x,.(y,z))
  i(.(x,y)) -> .(i(y),i(x))


-- SCC decomposition.

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

p1: .#(.(x,y),z) -> .#(x,.(y,z))
p2: .#(.(x,y),z) -> .#(y,z)
p3: i#(.(x,y)) -> .#(i(y),i(x))
p4: i#(.(x,y)) -> i#(y)
p5: i#(.(x,y)) -> i#(x)

and R consists of:

r1: .(|1|(),x) -> x
r2: .(x,|1|()) -> x
r3: .(i(x),x) -> |1|()
r4: .(x,i(x)) -> |1|()
r5: i(|1|()) -> |1|()
r6: i(i(x)) -> x
r7: .(i(y),.(y,z)) -> z
r8: .(y,.(i(y),z)) -> z
r9: .(.(x,y),z) -> .(x,.(y,z))
r10: i(.(x,y)) -> .(i(y),i(x))

The estimated dependency graph contains the following SCCs:

  {p4, p5}
  {p1, p2}


-- Reduction pair.

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

p1: i#(.(x,y)) -> i#(x)
p2: i#(.(x,y)) -> i#(y)

and R consists of:

r1: .(|1|(),x) -> x
r2: .(x,|1|()) -> x
r3: .(i(x),x) -> |1|()
r4: .(x,i(x)) -> |1|()
r5: i(|1|()) -> |1|()
r6: i(i(x)) -> x
r7: .(i(y),.(y,z)) -> z
r8: .(y,.(i(y),z)) -> z
r9: .(.(x,y),z) -> .(x,.(y,z))
r10: i(.(x,y)) -> .(i(y),i(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          i#_A(x1) = x1
          ._A(x1,x2) = x1 + x2 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          i#_A(x1) = 0
          ._A(x1,x2) = max{1, x1 - 5, x2 - 5}
          i#!_A(x1) = max{5, x1 - 2}
          .!_A(x1,x2) = max{x1 - 3, x2 + 4}
  
    partial status:
  
      pi(i#) = [1]
      pi(.) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

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

p1: .#(.(x,y),z) -> .#(x,.(y,z))
p2: .#(.(x,y),z) -> .#(y,z)

and R consists of:

r1: .(|1|(),x) -> x
r2: .(x,|1|()) -> x
r3: .(i(x),x) -> |1|()
r4: .(x,i(x)) -> |1|()
r5: i(|1|()) -> |1|()
r6: i(i(x)) -> x
r7: .(i(y),.(y,z)) -> z
r8: .(y,.(i(y),z)) -> z
r9: .(.(x,y),z) -> .(x,.(y,z))
r10: i(.(x,y)) -> .(i(y),i(x))

The set of usable rules consists of

  r1, r2, r3, r4, r7, r8, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          .#_A(x1,x2) = x1 + x2 + 4
          ._A(x1,x2) = x1 + x2 + 3
          |1|_A() = 1
          i_A(x1) = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          .#_A(x1,x2) = max{0, x1 - 4, x2 - 5}
          ._A(x1,x2) = max{3, x1 + 1, x2}
          |1|_A = 3
          i_A(x1) = max{1, x1 - 4}
          .#!_A(x1,x2) = max{2, x1 - 4, x2 - 8}
          .!_A(x1,x2) = max{1, x1 - 6, x2 - 11}
          |1|!_A = 0
          i!_A(x1) = max{1, x1 - 4}
  
    partial status:
  
      pi(.#) = []
      pi(.) = [1]
      pi(|1|) = []
      pi(i) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: .#(.(x,y),z) -> .#(x,.(y,z))

and R consists of:

r1: .(|1|(),x) -> x
r2: .(x,|1|()) -> x
r3: .(i(x),x) -> |1|()
r4: .(x,i(x)) -> |1|()
r5: i(|1|()) -> |1|()
r6: i(i(x)) -> x
r7: .(i(y),.(y,z)) -> z
r8: .(y,.(i(y),z)) -> z
r9: .(.(x,y),z) -> .(x,.(y,z))
r10: i(.(x,y)) -> .(i(y),i(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: .#(.(x,y),z) -> .#(x,.(y,z))

and R consists of:

r1: .(|1|(),x) -> x
r2: .(x,|1|()) -> x
r3: .(i(x),x) -> |1|()
r4: .(x,i(x)) -> |1|()
r5: i(|1|()) -> |1|()
r6: i(i(x)) -> x
r7: .(i(y),.(y,z)) -> z
r8: .(y,.(i(y),z)) -> z
r9: .(.(x,y),z) -> .(x,.(y,z))
r10: i(.(x,y)) -> .(i(y),i(x))

The set of usable rules consists of

  r1, r2, r3, r4, r7, r8, r9

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          .#_A(x1,x2) = x1 + x2 + 1
          ._A(x1,x2) = x1 + x2 + 3
          |1|_A() = 1
          i_A(x1) = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          .#_A(x1,x2) = max{0, x1 - 5, x2 - 6}
          ._A(x1,x2) = max{4, x1 + 1, x2}
          |1|_A = 4
          i_A(x1) = max{2, x1 - 5}
          .#!_A(x1,x2) = max{1, x1 - 5, x2 - 6}
          .!_A(x1,x2) = max{1, x1 - 6, x2 - 12}
          |1|!_A = 0
          i!_A(x1) = max{1, x1 - 6}
  
    partial status:
  
      pi(.#) = [1, 2]
      pi(.) = [1, 2]
      pi(|1|) = []
      pi(i) = []

The following rules are removed:

  p1

Then no dependency pair remains.