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:
  
      max/plus interpretations on natural numbers:
      
        sum2#_A(x1,x2) = x1 + 5
        cons_A(x1,x2) = max{x1, x2 + 1}
        +_A(x1,x2) = max{x1 + 6, x2}
        |0|_A = 4
        s_A(x1) = x1
        sum1_A(x1) = x1 + 7
        nil_A = 12
        sum2_A(x1,x2) = max{x1 + 10, x2 + 4}
        tests_A(x1) = x1 + 11
        true_A = 3
        and_A(x1,x2) = max{x1 - 26, x2 + 11}
        test_A(x1) = x1 + 8
        rands_A(x1,x2) = max{x1 + 19, x2 + 16}
        rand_A(x1) = x1 + 6
        done_A(x1) = x1 + 15
        eq_A(x1,x2) = max{x1 + 6, x2 + 4}
        f_A(x1) = x1 + 16
        g_A(x1) = x1 + 16
        |::|_A(x1,x2) = max{3, x1 - 9}
  
    precedence:
    
      sum2# = cons = + = |0| = sum1 = nil = sum2 = tests = true > |::| > s = and = test = rands = rand = done = eq = f = g
    
    partial status:
  
      pi(sum2#) = []
      pi(cons) = [1, 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) = [1]
      pi(eq) = [1, 2]
      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:
  
      max/plus interpretations on natural numbers:
      
        sum1#_A(x1) = x1 + 3
        cons_A(x1,x2) = max{x1 + 2, x2 + 2}
        +_A(x1,x2) = max{x1 + 6, x2}
        |0|_A = 47
        s_A(x1) = 44
        sum1_A(x1) = x1 + 31
        nil_A = 17
        sum2_A(x1,x2) = max{x1 + 9, x2 + 5}
        tests_A(x1) = 45
        true_A = 5
        and_A(x1,x2) = max{45, x1 + 2}
        test_A(x1) = x1 + 10
        rands_A(x1,x2) = max{33, x2 + 16}
        rand_A(x1) = max{90, x1 + 45}
        done_A(x1) = x1 + 3
        eq_A(x1,x2) = max{x1 + 6, x2 + 8}
        f_A(x1) = max{4, x1 + 2}
        g_A(x1) = max{2, x1 + 1}
        |::|_A(x1,x2) = max{15, x1 - 75}
  
    precedence:
    
      sum1# = cons = + = |0| = s = sum1 = nil = sum2 = tests = true = and = test = rands = rand = done = eq = f = g = |::|
    
    partial status:
  
      pi(sum1#) = []
      pi(cons) = [2]
      pi(+) = [2]
      pi(|0|) = []
      pi(s) = []
      pi(sum1) = []
      pi(nil) = []
      pi(sum2) = [1, 2]
      pi(tests) = []
      pi(true) = []
      pi(and) = []
      pi(test) = []
      pi(rands) = []
      pi(rand) = []
      pi(done) = [1]
      pi(eq) = [1, 2]
      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:
  
      max/plus interpretations on natural numbers:
      
        +#_A(x1,x2) = max{13, x1 + 1}
        s_A(x1) = max{12, x1}
        +_A(x1,x2) = max{x1 + 15, x2}
        |0|_A = 2
        sum1_A(x1) = x1 + 14
        nil_A = 0
        cons_A(x1,x2) = max{x1 + 15, x2 + 15}
        sum2_A(x1,x2) = max{x1 + 14, x2 + 14}
        tests_A(x1) = max{57, x1 + 27}
        true_A = 1
        and_A(x1,x2) = max{x1 + 1, x2 - 1}
        test_A(x1) = max{13, x1 + 4}
        rands_A(x1,x2) = max{x1 + 30, x2 + 40}
        rand_A(x1) = max{22, x1 + 1}
        done_A(x1) = x1 + 22
        eq_A(x1,x2) = max{23, x1 + 5, x2 + 5}
        f_A(x1) = x1 + 21
        g_A(x1) = x1 + 21
        |::|_A(x1,x2) = max{1, x1 - 20, x2}
  
    precedence:
    
      +# = + > s = |0| = nil > sum1 = cons = sum2 = tests = true = and = test = rands = rand = done = eq = f = g = |::|
    
    partial status:
  
      pi(+#) = [1]
      pi(s) = [1]
      pi(+) = [1, 2]
      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) = []
      pi(rand) = []
      pi(done) = [1]
      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:
  
      max/plus interpretations on natural numbers:
      
        rands#_A(x1,x2) = max{x1 + 42, x2 + 38}
        s_A(x1) = max{11, x1}
        |::|_A(x1,x2) = max{1, x1 - 34}
        rand_A(x1) = max{37, x1 + 12}
        |0|_A = 25
        +_A(x1,x2) = max{x1 + 10, x2}
        sum1_A(x1) = max{26, x1 + 2}
        nil_A = 1
        cons_A(x1,x2) = max{x1 + 8, x2 + 25}
        sum2_A(x1,x2) = max{x1 + 11, x2 + 8}
        tests_A(x1) = x1 + 93
        true_A = 1
        and_A(x1,x2) = max{x1 + 44, x2 + 93}
        test_A(x1) = 4
        rands_A(x1,x2) = max{12, x2 + 9}
        done_A(x1) = x1 + 9
        eq_A(x1,x2) = max{3, x1 + 2, x2 - 2}
        f_A(x1) = 1
        g_A(x1) = 5
  
    precedence:
    
      rands# > + = sum1 > s = |::| = rand = |0| = nil = cons = sum2 = tests = true = and = test = rands = done = eq = f = g
    
    partial status:
  
      pi(rands#) = [1]
      pi(s) = [1]
      pi(|::|) = []
      pi(rand) = []
      pi(|0|) = []
      pi(+) = [1, 2]
      pi(sum1) = [1]
      pi(nil) = []
      pi(cons) = [2]
      pi(sum2) = [1, 2]
      pi(tests) = [1]
      pi(true) = []
      pi(and) = []
      pi(test) = []
      pi(rands) = []
      pi(done) = []
      pi(eq) = [1]
      pi(f) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

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