YES

We show the termination of the relative TRS R/S:

  R:
  +(|0|(),y) -> y
  +(s(x),y) -> s(+(x,y))
  sum1(nil()) -> |0|()
  sum1(cons(x,y)) -> +(x,sum1(y))
  sum2(nil(),z) -> z
  sum2(cons(x,y),z) -> sum2(y,+(x,z))
  tests(|0|()) -> true()
  tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
  test(done(y)) -> eq(f(y),g(y))
  eq(x,x) -> true()
  rands(|0|(),y) -> done(y)
  rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))

  S:
  rand(x) -> x
  rand(x) -> rand(s(x))

-- SCC decomposition.

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

p1: +#(s(x),y) -> +#(x,y)
p2: sum1#(cons(x,y)) -> +#(x,sum1(y))
p3: sum1#(cons(x,y)) -> sum1#(y)
p4: sum2#(cons(x,y),z) -> sum2#(y,+(x,z))
p5: sum2#(cons(x,y),z) -> +#(x,z)
p6: tests#(s(x)) -> test#(rands(rand(|0|()),nil()))
p7: tests#(s(x)) -> rands#(rand(|0|()),nil())
p8: test#(done(y)) -> eq#(f(y),g(y))
p9: rands#(s(x),y) -> rands#(x,|::|(rand(|0|()),y))

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: sum1(nil()) -> |0|()
r4: sum1(cons(x,y)) -> +(x,sum1(y))
r5: sum2(nil(),z) -> z
r6: sum2(cons(x,y),z) -> sum2(y,+(x,z))
r7: tests(|0|()) -> true()
r8: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r9: test(done(y)) -> eq(f(y),g(y))
r10: eq(x,x) -> true()
r11: rands(|0|(),y) -> done(y)
r12: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p3}
  {p1}
  {p9}


-- Reduction pair.

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

p1: sum2#(cons(x,y),z) -> sum2#(y,+(x,z))

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: sum1(nil()) -> |0|()
r4: sum1(cons(x,y)) -> +(x,sum1(y))
r5: sum2(nil(),z) -> z
r6: sum2(cons(x,y),z) -> sum2(y,+(x,z))
r7: tests(|0|()) -> true()
r8: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r9: test(done(y)) -> eq(f(y),g(y))
r10: eq(x,x) -> true()
r11: rands(|0|(),y) -> done(y)
r12: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          sum2#_A(x1,x2) = x1 + x2 + 4
          cons_A(x1,x2) = x1 + x2 + 4
          +_A(x1,x2) = x1 + x2 + 1
          |0|_A() = 4
          s_A(x1) = x1
          sum1_A(x1) = x1 + 2
          nil_A() = 3
          sum2_A(x1,x2) = x1 + x2 + 4
          tests_A(x1) = x1 + 6
          true_A() = 1
          and_A(x1,x2) = 1
          test_A(x1) = 3
          rands_A(x1,x2) = x1
          rand_A(x1) = x1 + 1
          done_A(x1) = 3
          eq_A(x1,x2) = 2
          f_A(x1) = 1
          g_A(x1) = 1
          |::|_A(x1,x2) = 1
  
    precedence:
    
      cons = + = |0| > s = sum1 = nil = sum2 = true = and = test = eq = g > tests = rands = rand = done = f = |::| > sum2#
    
    partial status:
  
      pi(sum2#) = []
      pi(cons) = [2]
      pi(+) = [1, 2]
      pi(|0|) = []
      pi(s) = [1]
      pi(sum1) = [1]
      pi(nil) = []
      pi(sum2) = [1, 2]
      pi(tests) = [1]
      pi(true) = []
      pi(and) = []
      pi(test) = []
      pi(rands) = [1]
      pi(rand) = []
      pi(done) = []
      pi(eq) = []
      pi(f) = []
      pi(g) = []
      pi(|::|) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: sum1#(cons(x,y)) -> sum1#(y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: sum1(nil()) -> |0|()
r4: sum1(cons(x,y)) -> +(x,sum1(y))
r5: sum2(nil(),z) -> z
r6: sum2(cons(x,y),z) -> sum2(y,+(x,z))
r7: tests(|0|()) -> true()
r8: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r9: test(done(y)) -> eq(f(y),g(y))
r10: eq(x,x) -> true()
r11: rands(|0|(),y) -> done(y)
r12: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          sum1#_A(x1) = x1
          cons_A(x1,x2) = x1 + x2 + 3
          +_A(x1,x2) = x2 + 1
          |0|_A() = 0
          s_A(x1) = x1
          sum1_A(x1) = x1 + 1
          nil_A() = 1
          sum2_A(x1,x2) = x1 + x2 + 1
          tests_A(x1) = x1 + 3
          true_A() = 0
          and_A(x1,x2) = 1
          test_A(x1) = 2
          rands_A(x1,x2) = x1 + 2
          rand_A(x1) = x1 + 4
          done_A(x1) = 1
          eq_A(x1,x2) = 1
          f_A(x1) = 1
          g_A(x1) = 1
          |::|_A(x1,x2) = 1
  
    precedence:
    
      sum1# = cons = + > |0| = sum1 = nil = sum2 = true = and = test = rands = f > tests = rand = done = eq = g = |::| > s
    
    partial status:
  
      pi(sum1#) = []
      pi(cons) = [2]
      pi(+) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(sum1) = []
      pi(nil) = []
      pi(sum2) = []
      pi(tests) = [1]
      pi(true) = []
      pi(and) = []
      pi(test) = []
      pi(rands) = []
      pi(rand) = [1]
      pi(done) = []
      pi(eq) = []
      pi(f) = []
      pi(g) = []
      pi(|::|) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: +#(s(x),y) -> +#(x,y)

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: sum1(nil()) -> |0|()
r4: sum1(cons(x,y)) -> +(x,sum1(y))
r5: sum2(nil(),z) -> z
r6: sum2(cons(x,y),z) -> sum2(y,+(x,z))
r7: tests(|0|()) -> true()
r8: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r9: test(done(y)) -> eq(f(y),g(y))
r10: eq(x,x) -> true()
r11: rands(|0|(),y) -> done(y)
r12: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          +#_A(x1,x2) = x1
          s_A(x1) = x1
          +_A(x1,x2) = x1 + x2 + 2
          |0|_A() = 4
          sum1_A(x1) = x1 + 2
          nil_A() = 3
          cons_A(x1,x2) = x1 + x2 + 3
          sum2_A(x1,x2) = x1 + x2 + 2
          tests_A(x1) = x1 + 12
          true_A() = 1
          and_A(x1,x2) = 1
          test_A(x1) = 3
          rands_A(x1,x2) = x1 + 6
          rand_A(x1) = x1 + 1
          done_A(x1) = 3
          eq_A(x1,x2) = 2
          f_A(x1) = 1
          g_A(x1) = x1 + 4
          |::|_A(x1,x2) = 1
  
    precedence:
    
      + = |0| = sum1 > +# = s = nil = cons = sum2 = tests = true = test = rands = f > and = done = g = |::| > rand = eq
    
    partial status:
  
      pi(+#) = [1]
      pi(s) = [1]
      pi(+) = [1]
      pi(|0|) = []
      pi(sum1) = [1]
      pi(nil) = []
      pi(cons) = [2]
      pi(sum2) = [1, 2]
      pi(tests) = [1]
      pi(true) = []
      pi(and) = []
      pi(test) = []
      pi(rands) = [1]
      pi(rand) = []
      pi(done) = []
      pi(eq) = []
      pi(f) = []
      pi(g) = [1]
      pi(|::|) = []

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: rands#(s(x),y) -> rands#(x,|::|(rand(|0|()),y))

and R consists of:

r1: +(|0|(),y) -> y
r2: +(s(x),y) -> s(+(x,y))
r3: sum1(nil()) -> |0|()
r4: sum1(cons(x,y)) -> +(x,sum1(y))
r5: sum2(nil(),z) -> z
r6: sum2(cons(x,y),z) -> sum2(y,+(x,z))
r7: tests(|0|()) -> true()
r8: tests(s(x)) -> and(test(rands(rand(|0|()),nil())),x)
r9: test(done(y)) -> eq(f(y),g(y))
r10: eq(x,x) -> true()
r11: rands(|0|(),y) -> done(y)
r12: rands(s(x),y) -> rands(x,|::|(rand(|0|()),y))
r13: rand(x) -> x
r14: rand(x) -> rand(s(x))

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          rands#_A(x1,x2) = x1 + x2 + 6
          s_A(x1) = x1
          |::|_A(x1,x2) = x2
          rand_A(x1) = x1 + 1
          |0|_A() = 4
          +_A(x1,x2) = x1 + x2 + 1
          sum1_A(x1) = x1 + 3
          nil_A() = 3
          cons_A(x1,x2) = x1 + x2 + 2
          sum2_A(x1,x2) = x1 + x2 + 2
          tests_A(x1) = x1 + 2
          true_A() = 1
          and_A(x1,x2) = 1
          test_A(x1) = 3
          rands_A(x1,x2) = x1 + x2 + 1
          done_A(x1) = 3
          eq_A(x1,x2) = 2
          f_A(x1) = 1
          g_A(x1) = 1
  
    precedence:
    
      |::| = rand = |0| = sum1 = nil > + = cons = sum2 = true > tests = test = rands = done = eq = f > g > rands# > s = and
    
    partial status:
  
      pi(rands#) = [1]
      pi(s) = [1]
      pi(|::|) = []
      pi(rand) = []
      pi(|0|) = []
      pi(+) = [1]
      pi(sum1) = []
      pi(nil) = []
      pi(cons) = [2]
      pi(sum2) = []
      pi(tests) = [1]
      pi(true) = []
      pi(and) = []
      pi(test) = []
      pi(rands) = [1]
      pi(done) = []
      pi(eq) = []
      pi(f) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

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