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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sum2#_A(x1,x2) = x1
        cons_A(x1,x2) = x1 + x2 + 2
        +_A(x1,x2) = x2 + 1
        |0|_A() = 1
        s_A(x1) = x1
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        sum2_A(x1,x2) = x1 + x2
        tests_A(x1) = x1 + 1
        true_A() = 0
        and_A(x1,x2) = 0
        test_A(x1) = 2
        rands_A(x1,x2) = x1 + 1
        rand_A(x1) = x1 + 1
        done_A(x1) = 1
        eq_A(x1,x2) = 1
        f_A(x1) = x1 + 1
        g_A(x1) = x1
        |::|_A(x1,x2) = x2 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sum2#_A(x1,x2) = x1
        cons_A(x1,x2) = x1 + 1
        +_A(x1,x2) = 3
        |0|_A() = 3
        s_A(x1) = x1
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        sum2_A(x1,x2) = x1 + x2
        tests_A(x1) = 4
        true_A() = 3
        and_A(x1,x2) = 5
        test_A(x1) = 1
        rands_A(x1,x2) = x1
        rand_A(x1) = x1
        done_A(x1) = 4
        eq_A(x1,x2) = 2
        f_A(x1) = x1 + 1
        g_A(x1) = x1 + 1
        |::|_A(x1,x2) = x2 + 1
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sum1#_A(x1) = x1
        cons_A(x1,x2) = x1 + x2 + 2
        +_A(x1,x2) = x2 + 1
        |0|_A() = 1
        s_A(x1) = x1
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        sum2_A(x1,x2) = x1 + x2
        tests_A(x1) = x1 + 1
        true_A() = 0
        and_A(x1,x2) = 0
        test_A(x1) = 3
        rands_A(x1,x2) = x2 + 2
        rand_A(x1) = x1 + 1
        done_A(x1) = 1
        eq_A(x1,x2) = x1 + 1
        f_A(x1) = 1
        g_A(x1) = x1 + 1
        |::|_A(x1,x2) = x2
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        sum1#_A(x1) = 0
        cons_A(x1,x2) = x1 + 1
        +_A(x1,x2) = 3
        |0|_A() = 0
        s_A(x1) = 1
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        sum2_A(x1,x2) = x1
        tests_A(x1) = 0
        true_A() = 1
        and_A(x1,x2) = 1
        test_A(x1) = 3
        rands_A(x1,x2) = 2
        rand_A(x1) = 1
        done_A(x1) = 1
        eq_A(x1,x2) = 2
        f_A(x1) = 1
        g_A(x1) = x1 + 1
        |::|_A(x1,x2) = x2 + 1
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        s_A(x1) = x1
        +_A(x1,x2) = x1 + x2 + 1
        |0|_A() = 1
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        cons_A(x1,x2) = x1 + x2 + 2
        sum2_A(x1,x2) = x1 + x2
        tests_A(x1) = x1 + 1
        true_A() = 0
        and_A(x1,x2) = 0
        test_A(x1) = 3
        rands_A(x1,x2) = x1 + x2 + 1
        rand_A(x1) = x1 + 1
        done_A(x1) = 1
        eq_A(x1,x2) = x1 + 1
        f_A(x1) = 1
        g_A(x1) = x1 + 1
        |::|_A(x1,x2) = x2
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        +#_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
        +_A(x1,x2) = x1 + 1
        |0|_A() = 1
        sum1_A(x1) = 0
        nil_A() = 0
        cons_A(x1,x2) = x1 + x2
        sum2_A(x1,x2) = x1 + 1
        tests_A(x1) = 0
        true_A() = 3
        and_A(x1,x2) = 1
        test_A(x1) = 1
        rands_A(x1,x2) = x1 + 1
        rand_A(x1) = 1
        done_A(x1) = 1
        eq_A(x1,x2) = 2
        f_A(x1) = 1
        g_A(x1) = x1 + 1
        |::|_A(x1,x2) = x2 + 1
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        rands#_A(x1,x2) = x1
        s_A(x1) = x1
        |::|_A(x1,x2) = 1
        rand_A(x1) = x1 + 1
        |0|_A() = 1
        +_A(x1,x2) = x1 + x2 + 1
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        cons_A(x1,x2) = x1 + x2 + 2
        sum2_A(x1,x2) = x1 + x2
        tests_A(x1) = x1 + 1
        true_A() = 0
        and_A(x1,x2) = x2
        test_A(x1) = 3
        rands_A(x1,x2) = 2
        done_A(x1) = 1
        eq_A(x1,x2) = x1 + 1
        f_A(x1) = 1
        g_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        rands#_A(x1,x2) = x1
        s_A(x1) = x1 + 1
        |::|_A(x1,x2) = 1
        rand_A(x1) = 1
        |0|_A() = 1
        +_A(x1,x2) = x1 + x2 + 2
        sum1_A(x1) = x1 + 1
        nil_A() = 1
        cons_A(x1,x2) = x1 + x2 + 1
        sum2_A(x1,x2) = 1
        tests_A(x1) = 1
        true_A() = 3
        and_A(x1,x2) = x2
        test_A(x1) = 1
        rands_A(x1,x2) = 1
        done_A(x1) = 2
        eq_A(x1,x2) = 2
        f_A(x1) = 1
        g_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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