Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LEQ(s(X), s(Y)) → LEQ(X, Y)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(diff(p(X), Y)))
DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → DIFF(p(X), Y)
DIFF(X, Y) → P(X)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__s(X)) → S(X)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

LEQ(s(X), s(Y)) → LEQ(X, Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LEQ(s(X), s(Y)) → LEQ(X, Y)
  The graph contains the following edges 1 > 1, 2 > 2

(7) YES

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(diff(p(X), Y)))
0 → n__0
s(X) → n__s(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x1 / p(x1)] on the rule
DIFF(p(x1), x0)[ ]ⁿ[ ] → DIFF(p(x1), x0)[ ]ⁿ[x1 / p(x1)]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
DIFF(X, Y)[ ]ⁿ[ ] → DIFF(p(X), Y)[ ]ⁿ[ ]
    by OriginalRule from TRS P

(10) NO