YES

We show the termination of the TRS R:

  g(A()) -> A()
  g(B()) -> A()
  g(B()) -> B()
  g(C()) -> A()
  g(C()) -> B()
  g(C()) -> C()
  foldB(t,|0|()) -> t
  foldB(t,s(n)) -> f(foldB(t,n),B())
  foldC(t,|0|()) -> t
  foldC(t,s(n)) -> f(foldC(t,n),C())
  f(t,x) -> |f'|(t,g(x))
  |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
  |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
  |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
  |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

-- SCC decomposition.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: foldB#(t,s(n)) -> foldB#(t,n)
p3: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p4: foldC#(t,s(n)) -> foldC#(t,n)
p5: f#(t,x) -> |f'|#(t,g(x))
p6: f#(t,x) -> g#(x)
p7: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p8: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p9: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p10: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p4: foldB#(t,s(n)) -> foldB#(t,n)
p5: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p6: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p7: foldC#(t,s(n)) -> foldC#(t,n)
p8: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p9: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          foldB#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (8,8)
          s_A(x1) = x1 + (23,48)
          f#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (64,8)
          foldB_A(x1,x2) = x1 + ((0,1),(1,1)) x2 + (17,9)
          B_A() = (24,5)
          |f'|#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (37,8)
          g_A(x1) = ((0,1),(0,0)) x1 + (20,6)
          triple_A(x1,x2,x3) = ((0,1),(1,1)) x2 + ((0,1),(0,1)) x3 + (0,6)
          A_A() = (0,1)
          |0|_A() = (1,1)
          |f''|#_A(x1) = ((0,1),(0,0)) x1 + (26,8)
          foldC#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (24,8)
          foldC_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (1,1)
          C_A() = (26,6)
          |f''|_A(x1) = x1 + (3,2)
          |f'|_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (22,48)
          f_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (42,48)
  
    precedence:
    
      foldB = g = foldC = C = |f''| = |f'| = f > triple = A > foldB# = f# = |f'|# = |f''|# = foldC# > |0| > s = B
    
    partial status:
  
      pi(foldB#) = []
      pi(s) = [1]
      pi(f#) = []
      pi(foldB) = []
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(|0|) = []
      pi(|f''|#) = []
      pi(foldC#) = []
      pi(foldC) = []
      pi(C) = []
      pi(|f''|) = []
      pi(|f'|) = []
      pi(f) = []

The next rules are strictly ordered:

  p9

We remove them from the problem.

-- SCC decomposition.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p4: foldB#(t,s(n)) -> foldB#(t,n)
p5: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p6: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p7: foldC#(t,s(n)) -> foldC#(t,n)
p8: foldC#(t,s(n)) -> f#(foldC(t,n),C())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p4: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p5: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p6: foldC#(t,s(n)) -> foldC#(t,n)
p7: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p8: foldB#(t,s(n)) -> foldB#(t,n)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          foldB#_A(x1,x2) = ((1,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (13,17)
          s_A(x1) = x1 + (14,16)
          f#_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,0)) x2 + (21,19)
          foldB_A(x1,x2) = ((1,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (0,2)
          B_A() = (2,3)
          |f'|#_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,0)) x2 + (20,19)
          g_A(x1) = x1
          triple_A(x1,x2,x3) = ((0,0),(0,1)) x2 + ((0,0),(1,0)) x3 + (4,1)
          A_A() = (1,2)
          |f''|#_A(x1) = ((0,1),(0,1)) x1 + (20,17)
          |0|_A() = (0,0)
          foldC#_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (15,17)
          foldC_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (1,0)
          C_A() = (3,4)
          |f''|_A(x1) = ((0,1),(0,1)) x1 + (6,8)
          |f'|_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(1,1)) x2 + (10,7)
          f_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(1,1)) x2 + (11,7)
  
    precedence:
    
      foldB = g = triple > B = foldC = |f''| = |f'| = f > A = C > |0| > foldB# = f# = |f'|# = |f''|# = foldC# > s
    
    partial status:
  
      pi(foldB#) = []
      pi(s) = []
      pi(f#) = []
      pi(foldB) = [1]
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = [1]
      pi(triple) = []
      pi(A) = []
      pi(|f''|#) = []
      pi(|0|) = []
      pi(foldC#) = []
      pi(foldC) = []
      pi(C) = []
      pi(|f''|) = []
      pi(|f'|) = []
      pi(f) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p4: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p5: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p6: foldC#(t,s(n)) -> foldC#(t,n)
p7: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- Reduction pair.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p4: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p5: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p6: foldC#(t,s(n)) -> foldC#(t,n)
p7: foldC#(t,s(n)) -> f#(foldC(t,n),C())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          foldB#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (1,27)
          s_A(x1) = ((1,1),(0,1)) x1 + (11,26)
          f#_A(x1,x2) = ((0,1),(0,0)) x1 + (21,27)
          foldB_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,1)) x2 + (39,1)
          B_A() = (24,24)
          |f'|#_A(x1,x2) = ((0,1),(0,0)) x1 + (20,27)
          g_A(x1) = ((1,0),(1,0)) x1 + (2,1)
          triple_A(x1,x2,x3) = ((0,0),(1,1)) x2 + ((0,0),(0,1)) x3 + (2,1)
          A_A() = (21,22)
          |0|_A() = (1,1)
          |f''|#_A(x1) = ((0,1),(0,0)) x1 + (16,27)
          foldC#_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (12,27)
          foldC_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,1),(0,1)) x2 + (23,1)
          C_A() = (23,24)
          |f''|_A(x1) = ((0,1),(0,1)) x1 + (27,19)
          |f'|_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (22,2)
          f_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (24,3)
  
    precedence:
    
      |0| > |f''|# > foldB# = |f'|# > f# = foldC# > foldB = foldC = |f''| = |f'| = f > s > triple > B = g = A > C
    
    partial status:
  
      pi(foldB#) = []
      pi(s) = []
      pi(f#) = []
      pi(foldB) = []
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(|0|) = []
      pi(|f''|#) = []
      pi(foldC#) = []
      pi(foldC) = []
      pi(C) = []
      pi(|f''|) = []
      pi(|f'|) = []
      pi(f) = []

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p4: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p5: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p6: foldC#(t,s(n)) -> f#(foldC(t,n),C())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6}


-- Reduction pair.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p4: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p5: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p6: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          foldB#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (61,31)
          s_A(x1) = x1 + (62,2)
          f#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (64,31)
          foldB_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (63,30)
          B_A() = (9,31)
          |f'|#_A(x1,x2) = ((0,1),(0,0)) x1 + (63,31)
          g_A(x1) = ((1,0),(1,1)) x1 + (0,19)
          triple_A(x1,x2,x3) = ((0,0),(1,1)) x2 + ((0,0),(1,0)) x3 + (10,0)
          A_A() = (0,1)
          |f''|#_A(x1) = ((0,1),(0,0)) x1 + (11,31)
          |0|_A() = (2,0)
          foldC#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (7,31)
          foldC_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (2,3)
          C_A() = (11,1)
          |f''|_A(x1) = ((0,1),(0,1)) x1 + (15,5)
          |f'|_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (48,51)
          f_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (49,51)
  
    precedence:
    
      triple = |f'| > |f''| > foldC > f > foldB > |f''|# > |0| = foldC# > B = g = C > foldB# = s = f# = |f'|# > A
    
    partial status:
  
      pi(foldB#) = []
      pi(s) = []
      pi(f#) = []
      pi(foldB) = [1]
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(|f''|#) = []
      pi(|0|) = []
      pi(foldC#) = []
      pi(foldC) = []
      pi(C) = []
      pi(|f''|) = []
      pi(|f'|) = []
      pi(f) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p4: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p5: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p5}


-- Reduction pair.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          foldB#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (2,3)
          s_A(x1) = ((1,1),(0,0)) x1 + (28,4)
          f#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (27,3)
          foldB_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,1)) x2 + (35,5)
          B_A() = (1,2)
          |f'|#_A(x1,x2) = ((0,1),(0,0)) x1 + (26,3)
          g_A(x1) = ((0,1),(0,0)) x1 + (25,33)
          triple_A(x1,x2,x3) = ((0,0),(1,1)) x2 + ((0,1),(1,1)) x3 + (2,3)
          A_A() = (0,1)
          |0|_A() = (1,4)
          foldC_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,5)
          f_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(0,0)) x2 + (24,32)
          C_A() = (3,1)
          |f''|_A(x1) = ((0,1),(0,1)) x1 + (13,10)
          |f'|_A(x1,x2) = ((0,1),(0,1)) x1 + (24,32)
  
    precedence:
    
      foldB = foldC = f = |f''| = |f'| > s > triple > foldB# = |f'|# > f# = g = A > B > C > |0|
    
    partial status:
  
      pi(foldB#) = []
      pi(s) = []
      pi(f#) = []
      pi(foldB) = [1, 2]
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(|0|) = []
      pi(foldC) = []
      pi(f) = []
      pi(C) = []
      pi(|f''|) = []
      pi(|f'|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(t,x) -> |f'|#(t,g(x))
p2: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

  (no SCCs)