YES

We show the termination of the TRS R:

  a__U11(tt(),V2) -> a__U12(a__isNat(V2))
  a__U12(tt()) -> tt()
  a__U21(tt()) -> tt()
  a__U31(tt(),N) -> mark(N)
  a__U41(tt(),M,N) -> a__U42(a__isNat(N),M,N)
  a__U42(tt(),M,N) -> s(a__plus(mark(N),mark(M)))
  a__isNat(|0|()) -> tt()
  a__isNat(plus(V1,V2)) -> a__U11(a__isNat(V1),V2)
  a__isNat(s(V1)) -> a__U21(a__isNat(V1))
  a__plus(N,|0|()) -> a__U31(a__isNat(N),N)
  a__plus(N,s(M)) -> a__U41(a__isNat(M),M,N)
  mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
  mark(U12(X)) -> a__U12(mark(X))
  mark(isNat(X)) -> a__isNat(X)
  mark(U21(X)) -> a__U21(mark(X))
  mark(U31(X1,X2)) -> a__U31(mark(X1),X2)
  mark(U41(X1,X2,X3)) -> a__U41(mark(X1),X2,X3)
  mark(U42(X1,X2,X3)) -> a__U42(mark(X1),X2,X3)
  mark(plus(X1,X2)) -> a__plus(mark(X1),mark(X2))
  mark(tt()) -> tt()
  mark(s(X)) -> s(mark(X))
  mark(|0|()) -> |0|()
  a__U11(X1,X2) -> U11(X1,X2)
  a__U12(X) -> U12(X)
  a__isNat(X) -> isNat(X)
  a__U21(X) -> U21(X)
  a__U31(X1,X2) -> U31(X1,X2)
  a__U41(X1,X2,X3) -> U41(X1,X2,X3)
  a__U42(X1,X2,X3) -> U42(X1,X2,X3)
  a__plus(X1,X2) -> plus(X1,X2)

-- SCC decomposition.

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

p1: a__U11#(tt(),V2) -> a__U12#(a__isNat(V2))
p2: a__U11#(tt(),V2) -> a__isNat#(V2)
p3: a__U31#(tt(),N) -> mark#(N)
p4: a__U41#(tt(),M,N) -> a__U42#(a__isNat(N),M,N)
p5: a__U41#(tt(),M,N) -> a__isNat#(N)
p6: a__U42#(tt(),M,N) -> a__plus#(mark(N),mark(M))
p7: a__U42#(tt(),M,N) -> mark#(N)
p8: a__U42#(tt(),M,N) -> mark#(M)
p9: a__isNat#(plus(V1,V2)) -> a__U11#(a__isNat(V1),V2)
p10: a__isNat#(plus(V1,V2)) -> a__isNat#(V1)
p11: a__isNat#(s(V1)) -> a__U21#(a__isNat(V1))
p12: a__isNat#(s(V1)) -> a__isNat#(V1)
p13: a__plus#(N,|0|()) -> a__U31#(a__isNat(N),N)
p14: a__plus#(N,|0|()) -> a__isNat#(N)
p15: a__plus#(N,s(M)) -> a__U41#(a__isNat(M),M,N)
p16: a__plus#(N,s(M)) -> a__isNat#(M)
p17: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2)
p18: mark#(U11(X1,X2)) -> mark#(X1)
p19: mark#(U12(X)) -> a__U12#(mark(X))
p20: mark#(U12(X)) -> mark#(X)
p21: mark#(isNat(X)) -> a__isNat#(X)
p22: mark#(U21(X)) -> a__U21#(mark(X))
p23: mark#(U21(X)) -> mark#(X)
p24: mark#(U31(X1,X2)) -> a__U31#(mark(X1),X2)
p25: mark#(U31(X1,X2)) -> mark#(X1)
p26: mark#(U41(X1,X2,X3)) -> a__U41#(mark(X1),X2,X3)
p27: mark#(U41(X1,X2,X3)) -> mark#(X1)
p28: mark#(U42(X1,X2,X3)) -> a__U42#(mark(X1),X2,X3)
p29: mark#(U42(X1,X2,X3)) -> mark#(X1)
p30: mark#(plus(X1,X2)) -> a__plus#(mark(X1),mark(X2))
p31: mark#(plus(X1,X2)) -> mark#(X1)
p32: mark#(plus(X1,X2)) -> mark#(X2)
p33: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__U11(tt(),V2) -> a__U12(a__isNat(V2))
r2: a__U12(tt()) -> tt()
r3: a__U21(tt()) -> tt()
r4: a__U31(tt(),N) -> mark(N)
r5: a__U41(tt(),M,N) -> a__U42(a__isNat(N),M,N)
r6: a__U42(tt(),M,N) -> s(a__plus(mark(N),mark(M)))
r7: a__isNat(|0|()) -> tt()
r8: a__isNat(plus(V1,V2)) -> a__U11(a__isNat(V1),V2)
r9: a__isNat(s(V1)) -> a__U21(a__isNat(V1))
r10: a__plus(N,|0|()) -> a__U31(a__isNat(N),N)
r11: a__plus(N,s(M)) -> a__U41(a__isNat(M),M,N)
r12: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r13: mark(U12(X)) -> a__U12(mark(X))
r14: mark(isNat(X)) -> a__isNat(X)
r15: mark(U21(X)) -> a__U21(mark(X))
r16: mark(U31(X1,X2)) -> a__U31(mark(X1),X2)
r17: mark(U41(X1,X2,X3)) -> a__U41(mark(X1),X2,X3)
r18: mark(U42(X1,X2,X3)) -> a__U42(mark(X1),X2,X3)
r19: mark(plus(X1,X2)) -> a__plus(mark(X1),mark(X2))
r20: mark(tt()) -> tt()
r21: mark(s(X)) -> s(mark(X))
r22: mark(|0|()) -> |0|()
r23: a__U11(X1,X2) -> U11(X1,X2)
r24: a__U12(X) -> U12(X)
r25: a__isNat(X) -> isNat(X)
r26: a__U21(X) -> U21(X)
r27: a__U31(X1,X2) -> U31(X1,X2)
r28: a__U41(X1,X2,X3) -> U41(X1,X2,X3)
r29: a__U42(X1,X2,X3) -> U42(X1,X2,X3)
r30: a__plus(X1,X2) -> plus(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p3, p4, p6, p7, p8, p13, p15, p18, p20, p23, p24, p25, p26, p27, p28, p29, p30, p31, p32, p33}
  {p2, p9, p10, p12}


-- Reduction pair.

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

p1: a__U42#(tt(),M,N) -> mark#(M)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(plus(X1,X2)) -> mark#(X2)
p4: mark#(plus(X1,X2)) -> mark#(X1)
p5: mark#(plus(X1,X2)) -> a__plus#(mark(X1),mark(X2))
p6: a__plus#(N,s(M)) -> a__U41#(a__isNat(M),M,N)
p7: a__U41#(tt(),M,N) -> a__U42#(a__isNat(N),M,N)
p8: a__U42#(tt(),M,N) -> mark#(N)
p9: mark#(U42(X1,X2,X3)) -> mark#(X1)
p10: mark#(U42(X1,X2,X3)) -> a__U42#(mark(X1),X2,X3)
p11: a__U42#(tt(),M,N) -> a__plus#(mark(N),mark(M))
p12: a__plus#(N,|0|()) -> a__U31#(a__isNat(N),N)
p13: a__U31#(tt(),N) -> mark#(N)
p14: mark#(U41(X1,X2,X3)) -> mark#(X1)
p15: mark#(U41(X1,X2,X3)) -> a__U41#(mark(X1),X2,X3)
p16: mark#(U31(X1,X2)) -> mark#(X1)
p17: mark#(U31(X1,X2)) -> a__U31#(mark(X1),X2)
p18: mark#(U21(X)) -> mark#(X)
p19: mark#(U12(X)) -> mark#(X)
p20: mark#(U11(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__U11(tt(),V2) -> a__U12(a__isNat(V2))
r2: a__U12(tt()) -> tt()
r3: a__U21(tt()) -> tt()
r4: a__U31(tt(),N) -> mark(N)
r5: a__U41(tt(),M,N) -> a__U42(a__isNat(N),M,N)
r6: a__U42(tt(),M,N) -> s(a__plus(mark(N),mark(M)))
r7: a__isNat(|0|()) -> tt()
r8: a__isNat(plus(V1,V2)) -> a__U11(a__isNat(V1),V2)
r9: a__isNat(s(V1)) -> a__U21(a__isNat(V1))
r10: a__plus(N,|0|()) -> a__U31(a__isNat(N),N)
r11: a__plus(N,s(M)) -> a__U41(a__isNat(M),M,N)
r12: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r13: mark(U12(X)) -> a__U12(mark(X))
r14: mark(isNat(X)) -> a__isNat(X)
r15: mark(U21(X)) -> a__U21(mark(X))
r16: mark(U31(X1,X2)) -> a__U31(mark(X1),X2)
r17: mark(U41(X1,X2,X3)) -> a__U41(mark(X1),X2,X3)
r18: mark(U42(X1,X2,X3)) -> a__U42(mark(X1),X2,X3)
r19: mark(plus(X1,X2)) -> a__plus(mark(X1),mark(X2))
r20: mark(tt()) -> tt()
r21: mark(s(X)) -> s(mark(X))
r22: mark(|0|()) -> |0|()
r23: a__U11(X1,X2) -> U11(X1,X2)
r24: a__U12(X) -> U12(X)
r25: a__isNat(X) -> isNat(X)
r26: a__U21(X) -> U21(X)
r27: a__U31(X1,X2) -> U31(X1,X2)
r28: a__U41(X1,X2,X3) -> U41(X1,X2,X3)
r29: a__U42(X1,X2,X3) -> U42(X1,X2,X3)
r30: a__plus(X1,X2) -> plus(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      a__U42#_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,0,0),(0,0,0,0)) x2 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(1,0,0,0)) x3 + (8,2,2,36)
      tt_A() = (17,1,73,42)
      mark#_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (0,21,92,53)
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (3,1,1,0)
      plus_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (25,1,0,2)
      a__plus#_A(x1,x2) = x1 + ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x2 + (24,36,94,77)
      mark_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (0,31,58,0)
      a__U41#_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,0)) x1 + x2 + x3 + (9,0,24,59)
      a__isNat_A(x1) = (17,22,56,41)
      U42_A(x1,x2,x3) = x1 + x2 + x3 + (11,2,0,4)
      |0|_A() = (29,1,0,0)
      a__U31#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,1,1,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + (0,26,0,0)
      U41_A(x1,x2,x3) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x3 + (11,1,0,3)
      U31_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + (37,15,0,1)
      U21_A(x1) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,1,0)) x1 + (0,0,0,26)
      U12_A(x1) = x1 + (0,1,0,1)
      U11_A(x1,x2) = x1 + (0,22,1,1)
      a__U11_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (0,22,55,23)
      a__U12_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (0,22,0,0)
      a__U21_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (0,2,57,25)
      a__U31_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + (37,15,15,0)
      a__U41_A(x1,x2,x3) = x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + x3 + (11,11,28,2)
      a__U42_A(x1,x2,x3) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x2 + x3 + (11,2,10,3)
      a__plus_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x2 + (25,33,26,1)
      isNat_A(x1) = (17,0,0,42)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(U21(X)) -> mark#(X)
p2: mark#(U12(X)) -> mark#(X)
p3: mark#(U11(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__U11(tt(),V2) -> a__U12(a__isNat(V2))
r2: a__U12(tt()) -> tt()
r3: a__U21(tt()) -> tt()
r4: a__U31(tt(),N) -> mark(N)
r5: a__U41(tt(),M,N) -> a__U42(a__isNat(N),M,N)
r6: a__U42(tt(),M,N) -> s(a__plus(mark(N),mark(M)))
r7: a__isNat(|0|()) -> tt()
r8: a__isNat(plus(V1,V2)) -> a__U11(a__isNat(V1),V2)
r9: a__isNat(s(V1)) -> a__U21(a__isNat(V1))
r10: a__plus(N,|0|()) -> a__U31(a__isNat(N),N)
r11: a__plus(N,s(M)) -> a__U41(a__isNat(M),M,N)
r12: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r13: mark(U12(X)) -> a__U12(mark(X))
r14: mark(isNat(X)) -> a__isNat(X)
r15: mark(U21(X)) -> a__U21(mark(X))
r16: mark(U31(X1,X2)) -> a__U31(mark(X1),X2)
r17: mark(U41(X1,X2,X3)) -> a__U41(mark(X1),X2,X3)
r18: mark(U42(X1,X2,X3)) -> a__U42(mark(X1),X2,X3)
r19: mark(plus(X1,X2)) -> a__plus(mark(X1),mark(X2))
r20: mark(tt()) -> tt()
r21: mark(s(X)) -> s(mark(X))
r22: mark(|0|()) -> |0|()
r23: a__U11(X1,X2) -> U11(X1,X2)
r24: a__U12(X) -> U12(X)
r25: a__isNat(X) -> isNat(X)
r26: a__U21(X) -> U21(X)
r27: a__U31(X1,X2) -> U31(X1,X2)
r28: a__U41(X1,X2,X3) -> U41(X1,X2,X3)
r29: a__U42(X1,X2,X3) -> U42(X1,X2,X3)
r30: a__plus(X1,X2) -> plus(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: mark#(U21(X)) -> mark#(X)
p2: mark#(U11(X1,X2)) -> mark#(X1)
p3: mark#(U12(X)) -> mark#(X)

and R consists of:

r1: a__U11(tt(),V2) -> a__U12(a__isNat(V2))
r2: a__U12(tt()) -> tt()
r3: a__U21(tt()) -> tt()
r4: a__U31(tt(),N) -> mark(N)
r5: a__U41(tt(),M,N) -> a__U42(a__isNat(N),M,N)
r6: a__U42(tt(),M,N) -> s(a__plus(mark(N),mark(M)))
r7: a__isNat(|0|()) -> tt()
r8: a__isNat(plus(V1,V2)) -> a__U11(a__isNat(V1),V2)
r9: a__isNat(s(V1)) -> a__U21(a__isNat(V1))
r10: a__plus(N,|0|()) -> a__U31(a__isNat(N),N)
r11: a__plus(N,s(M)) -> a__U41(a__isNat(M),M,N)
r12: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r13: mark(U12(X)) -> a__U12(mark(X))
r14: mark(isNat(X)) -> a__isNat(X)
r15: mark(U21(X)) -> a__U21(mark(X))
r16: mark(U31(X1,X2)) -> a__U31(mark(X1),X2)
r17: mark(U41(X1,X2,X3)) -> a__U41(mark(X1),X2,X3)
r18: mark(U42(X1,X2,X3)) -> a__U42(mark(X1),X2,X3)
r19: mark(plus(X1,X2)) -> a__plus(mark(X1),mark(X2))
r20: mark(tt()) -> tt()
r21: mark(s(X)) -> s(mark(X))
r22: mark(|0|()) -> |0|()
r23: a__U11(X1,X2) -> U11(X1,X2)
r24: a__U12(X) -> U12(X)
r25: a__isNat(X) -> isNat(X)
r26: a__U21(X) -> U21(X)
r27: a__U31(X1,X2) -> U31(X1,X2)
r28: a__U41(X1,X2,X3) -> U41(X1,X2,X3)
r29: a__U42(X1,X2,X3) -> U42(X1,X2,X3)
r30: a__plus(X1,X2) -> plus(X1,X2)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      mark#_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,0,0,1)) x1
      U21_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (0,1,0,0)
      U11_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,1,1)) x1 + ((0,0,0,0),(1,0,0,0),(1,0,0,0),(1,1,0,0)) x2 + (1,1,1,1)
      U12_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (0,1,1,1)

The next rules are strictly ordered:

  p1, p2, p3

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

-- Reduction pair.

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

p1: a__U11#(tt(),V2) -> a__isNat#(V2)
p2: a__isNat#(s(V1)) -> a__isNat#(V1)
p3: a__isNat#(plus(V1,V2)) -> a__isNat#(V1)
p4: a__isNat#(plus(V1,V2)) -> a__U11#(a__isNat(V1),V2)

and R consists of:

r1: a__U11(tt(),V2) -> a__U12(a__isNat(V2))
r2: a__U12(tt()) -> tt()
r3: a__U21(tt()) -> tt()
r4: a__U31(tt(),N) -> mark(N)
r5: a__U41(tt(),M,N) -> a__U42(a__isNat(N),M,N)
r6: a__U42(tt(),M,N) -> s(a__plus(mark(N),mark(M)))
r7: a__isNat(|0|()) -> tt()
r8: a__isNat(plus(V1,V2)) -> a__U11(a__isNat(V1),V2)
r9: a__isNat(s(V1)) -> a__U21(a__isNat(V1))
r10: a__plus(N,|0|()) -> a__U31(a__isNat(N),N)
r11: a__plus(N,s(M)) -> a__U41(a__isNat(M),M,N)
r12: mark(U11(X1,X2)) -> a__U11(mark(X1),X2)
r13: mark(U12(X)) -> a__U12(mark(X))
r14: mark(isNat(X)) -> a__isNat(X)
r15: mark(U21(X)) -> a__U21(mark(X))
r16: mark(U31(X1,X2)) -> a__U31(mark(X1),X2)
r17: mark(U41(X1,X2,X3)) -> a__U41(mark(X1),X2,X3)
r18: mark(U42(X1,X2,X3)) -> a__U42(mark(X1),X2,X3)
r19: mark(plus(X1,X2)) -> a__plus(mark(X1),mark(X2))
r20: mark(tt()) -> tt()
r21: mark(s(X)) -> s(mark(X))
r22: mark(|0|()) -> |0|()
r23: a__U11(X1,X2) -> U11(X1,X2)
r24: a__U12(X) -> U12(X)
r25: a__isNat(X) -> isNat(X)
r26: a__U21(X) -> U21(X)
r27: a__U31(X1,X2) -> U31(X1,X2)
r28: a__U41(X1,X2,X3) -> U41(X1,X2,X3)
r29: a__U42(X1,X2,X3) -> U42(X1,X2,X3)
r30: a__plus(X1,X2) -> plus(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r7, r8, r9, r23, r24, r25, r26

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      a__U11#_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x2 + (0,0,0,1)
      tt_A() = (1,7,8,3)
      a__isNat#_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x1
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + (2,1,1,0)
      plus_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(0,0,1,0),(1,1,1,1)) x2 + (0,2,1,0)
      a__isNat_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (4,6,7,1)
      a__U12_A(x1) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (2,0,1,2)
      U12_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,1,0,3)
      a__U11_A(x1,x2) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + x2 + (3,5,6,0)
      a__U21_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,1)) x1 + (1,1,1,0)
      U11_A(x1,x2) = (0,6,7,1)
      U21_A(x1) = (0,2,2,1)
      |0|_A() = (1,1,1,0)
      isNat_A(x1) = (0,0,0,2)

The next rules are strictly ordered:

  p1, p2, p3, p4

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