Term Rewriting System R:
[x, y, z]
times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TIMES(x, plus(y, s(z))) -> PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) -> PLUS(y, times(s(z), 0))
TIMES(x, plus(y, s(z))) -> TIMES(s(z), 0)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, s(y)) -> PLUS(times(x, y), x)
TIMES(x, s(y)) -> TIMES(x, y)
PLUS(x, s(y)) -> PLUS(x, y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pair:

PLUS(x, s(y)) -> PLUS(x, y)


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





The following dependency pair can be strictly oriented:

PLUS(x, s(y)) -> PLUS(x, y)


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Remaining


Dependency Pair:


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

TIMES(x, s(y)) -> TIMES(x, y)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))




Termination of R could not be shown.
Duration:
0:01 minutes