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. lexicographic path order with precedence:
    
      precedence:
      
        |::| > test > g > f > eq > rands > done > rand > and > sum2 > sum1 > |0| > true > nil > tests > s > + > sum2# > cons
      
      argument filter:
    
        pi(sum2#) = [1]
        pi(cons) = [1, 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) = 2
        pi(test) = [1]
        pi(rands) = [1]
        pi(rand) = 1
        pi(done) = []
        pi(eq) = [2]
        pi(f) = [1]
        pi(g) = []
        pi(|::|) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > true > eq > done > rand > |0| > rands > test > and > sum2# > nil > tests > + > sum2 > sum1 > s > cons
      
      argument filter:
    
        pi(sum2#) = [1]
        pi(cons) = 2
        pi(+) = [1]
        pi(|0|) = []
        pi(s) = 1
        pi(sum1) = [1]
        pi(nil) = []
        pi(sum2) = [1, 2]
        pi(tests) = [1]
        pi(true) = []
        pi(and) = 2
        pi(test) = [1]
        pi(rands) = []
        pi(rand) = 1
        pi(done) = []
        pi(eq) = [2]
        pi(f) = 1
        pi(g) = []
        pi(|::|) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > true > eq > done > rand > |0| > rands > test > and > nil > tests > + > sum2 > sum1 > s > cons > sum2#
      
      argument filter:
    
        pi(sum2#) = []
        pi(cons) = [2]
        pi(+) = [1]
        pi(|0|) = []
        pi(s) = 1
        pi(sum1) = 1
        pi(nil) = []
        pi(sum2) = 1
        pi(tests) = [1]
        pi(true) = []
        pi(and) = 2
        pi(test) = [1]
        pi(rands) = []
        pi(rand) = [1]
        pi(done) = []
        pi(eq) = 2
        pi(f) = 1
        pi(g) = []
        pi(|::|) = [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. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > test > eq > done > rand > rands > and > sum2 > sum1 > |0| > true > nil > tests > + > s > cons > sum1#
      
      argument filter:
    
        pi(sum1#) = 1
        pi(cons) = 2
        pi(+) = 2
        pi(|0|) = []
        pi(s) = 1
        pi(sum1) = []
        pi(nil) = []
        pi(sum2) = 2
        pi(tests) = 1
        pi(true) = []
        pi(and) = 2
        pi(test) = [1]
        pi(rands) = 2
        pi(rand) = 1
        pi(done) = 1
        pi(eq) = [1, 2]
        pi(f) = 1
        pi(g) = 1
        pi(|::|) = 2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > true > eq > done > rand > |0| > rands > test > and > nil > tests > + > sum2 > sum1 > s > cons > sum1#
      
      argument filter:
    
        pi(sum1#) = 1
        pi(cons) = 2
        pi(+) = 2
        pi(|0|) = []
        pi(s) = 1
        pi(sum1) = []
        pi(nil) = []
        pi(sum2) = [2]
        pi(tests) = [1]
        pi(true) = []
        pi(and) = 2
        pi(test) = [1]
        pi(rands) = [2]
        pi(rand) = 1
        pi(done) = 1
        pi(eq) = [1, 2]
        pi(f) = 1
        pi(g) = [1]
        pi(|::|) = 2
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > true > eq > done > rand > |0| > rands > test > and > nil > tests > + > sum2 > sum1 > s > cons > sum1#
      
      argument filter:
    
        pi(sum1#) = [1]
        pi(cons) = [2]
        pi(+) = 2
        pi(|0|) = []
        pi(s) = 1
        pi(sum1) = []
        pi(nil) = []
        pi(sum2) = []
        pi(tests) = [1]
        pi(true) = []
        pi(and) = []
        pi(test) = [1]
        pi(rands) = [2]
        pi(rand) = 1
        pi(done) = 1
        pi(eq) = [1, 2]
        pi(f) = 1
        pi(g) = 1
        pi(|::|) = 2
    

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. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > sum1 > |0| > tests > test > nil > done > eq > rand > rands > and > true > sum2 > + > cons > s > +#
      
      argument filter:
    
        pi(+#) = [1, 2]
        pi(s) = 1
        pi(+) = [1, 2]
        pi(|0|) = []
        pi(sum1) = [1]
        pi(nil) = []
        pi(cons) = [1, 2]
        pi(sum2) = [1, 2]
        pi(tests) = []
        pi(true) = []
        pi(and) = 1
        pi(test) = [1]
        pi(rands) = 2
        pi(rand) = [1]
        pi(done) = 1
        pi(eq) = [1, 2]
        pi(f) = 1
        pi(g) = 1
        pi(|::|) = 2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > true > eq > done > rand > |0| > rands > test > and > nil > tests > + > sum2 > sum1 > cons > +# > s
      
      argument filter:
    
        pi(+#) = [1, 2]
        pi(s) = 1
        pi(+) = 1
        pi(|0|) = []
        pi(sum1) = 1
        pi(nil) = []
        pi(cons) = [1, 2]
        pi(sum2) = 1
        pi(tests) = []
        pi(true) = []
        pi(and) = 1
        pi(test) = 1
        pi(rands) = 2
        pi(rand) = []
        pi(done) = 1
        pi(eq) = [1, 2]
        pi(f) = 1
        pi(g) = 1
        pi(|::|) = 2
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        |::| > g > f > |0| > true > eq > done > rand > rands > tests > test > and > nil > + > sum2 > sum1 > cons > s > +#
      
      argument filter:
    
        pi(+#) = [1]
        pi(s) = [1]
        pi(+) = 1
        pi(|0|) = []
        pi(sum1) = 1
        pi(nil) = []
        pi(cons) = [2]
        pi(sum2) = 1
        pi(tests) = []
        pi(true) = []
        pi(and) = 1
        pi(test) = []
        pi(rands) = [2]
        pi(rand) = []
        pi(done) = 1
        pi(eq) = [1, 2]
        pi(f) = 1
        pi(g) = 1
        pi(|::|) = 2
    

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. lexicographic path order with precedence:
    
      precedence:
      
        g > f > test > eq > rands# > rand > |::| > rands > and > sum1 > |0| > done > true > nil > tests > sum2 > cons > + > s
      
      argument filter:
    
        pi(rands#) = [1]
        pi(s) = 1
        pi(|::|) = 2
        pi(rand) = [1]
        pi(|0|) = []
        pi(+) = [1, 2]
        pi(sum1) = [1]
        pi(nil) = []
        pi(cons) = [1, 2]
        pi(sum2) = [1, 2]
        pi(tests) = 1
        pi(true) = []
        pi(and) = 2
        pi(test) = [1]
        pi(rands) = 2
        pi(done) = 1
        pi(eq) = []
        pi(f) = [1]
        pi(g) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g > f > true > eq > rand > |::| > rands > done > |0| > tests > and > s > + > test > nil > sum2 > sum1 > cons > rands#
      
      argument filter:
    
        pi(rands#) = 1
        pi(s) = 1
        pi(|::|) = 2
        pi(rand) = []
        pi(|0|) = []
        pi(+) = 1
        pi(sum1) = 1
        pi(nil) = []
        pi(cons) = 1
        pi(sum2) = [1, 2]
        pi(tests) = []
        pi(true) = []
        pi(and) = []
        pi(test) = [1]
        pi(rands) = []
        pi(done) = []
        pi(eq) = []
        pi(f) = 1
        pi(g) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        g > f > true > eq > done > |0| > rand > cons > s > |::| > rands > test > and > nil > tests > + > sum2 > sum1 > rands#
      
      argument filter:
    
        pi(rands#) = 1
        pi(s) = [1]
        pi(|::|) = []
        pi(rand) = []
        pi(|0|) = []
        pi(+) = 1
        pi(sum1) = [1]
        pi(nil) = []
        pi(cons) = [1]
        pi(sum2) = 2
        pi(tests) = []
        pi(true) = []
        pi(and) = []
        pi(test) = [1]
        pi(rands) = []
        pi(done) = []
        pi(eq) = []
        pi(f) = 1
        pi(g) = 1
    

The next rules are strictly ordered:

  p1

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