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

  r1, r2, r3, r4, r5, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        i#_A(x1) = x1 + 1
        ._A(x1,x2) = max{11, x1 + 4, x2 + 2}
        |1|_A = 3
        i_A(x1) = max{5, x1 - 5}
  
    base order 2:
  
      precedence:
      
        . > i# = i > |1|
      
  
    partial status:
  
      pi(i#) = [1]
      pi(.) = []
      pi(|1|) = []
      pi(i) = []

The following rules are removed:

  p2



-- SCC decomposition.

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

p1: 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:

  {p1}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  r2, r3, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        i#_A(x1) = x1 + 6
        ._A(x1,x2) = max{x1 + 7, x2 - 4}
        |1|_A = 1
        i_A(x1) = max{2, x1 - 10}
  
    base order 2:
  
      precedence:
      
        . > i# = i > |1|
      
  
    partial status:
  
      pi(i#) = [1]
      pi(.) = [1]
      pi(|1|) = []
      pi(i) = []

The following rules are removed:

  p1

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, r5, r7, r8

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        .#_A(x1,x2) = max{4, x1 + 2}
        ._A(x1,x2) = max{11, x1 + 3, x2 + 3}
        |1|_A = 5
        i_A(x1) = max{6, x1 - 4}
  
    base order 2:
  
      precedence:
      
        . = i > .# = |1|
      
  
    partial status:
  
      pi(.#) = [1]
      pi(.) = [2]
      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

  r2, r3, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        .#_A(x1,x2) = x1 + 20
        ._A(x1,x2) = max{7, x1 + 4, x2 - 8}
        |1|_A = 5
        i_A(x1) = max{6, x1 - 10}
  
    base order 2:
  
      precedence:
      
        .# > . > i > |1|
      
  
    partial status:
  
      pi(.#) = []
      pi(.) = [1]
      pi(|1|) = []
      pi(i) = []

The following rules are removed:

  p1

Then no dependency pair remains.