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

(0) Obligation:

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

or(true, y) → true
or(x, true) → true
or(false, false) → false
mem(x, nil) → false
mem(x, set(y)) → =(x, y)
mem(x, union(y, z)) → or(mem(x, y), mem(x, z))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

or(true, y) → true
or(x, true) → true
or(false, false) → false
mem(x, nil) → false
mem(x, set(y)) → =(x, y)
mem(x, union(y, z)) → or(mem(x, y), mem(x, z))

The set Q consists of the following terms:

or(true, x0)
or(x0, true)
or(false, false)
mem(x0, nil)
mem(x0, set(x1))
mem(x0, union(x1, x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

MEM(x, union(y, z)) → OR(mem(x, y), mem(x, z))
MEM(x, union(y, z)) → MEM(x, y)
MEM(x, union(y, z)) → MEM(x, z)

The TRS R consists of the following rules:

or(true, y) → true
or(x, true) → true
or(false, false) → false
mem(x, nil) → false
mem(x, set(y)) → =(x, y)
mem(x, union(y, z)) → or(mem(x, y), mem(x, z))

The set Q consists of the following terms:

or(true, x0)
or(x0, true)
or(false, false)
mem(x0, nil)
mem(x0, set(x1))
mem(x0, union(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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

MEM(x, union(y, z)) → MEM(x, z)
MEM(x, union(y, z)) → MEM(x, y)

The TRS R consists of the following rules:

or(true, y) → true
or(x, true) → true
or(false, false) → false
mem(x, nil) → false
mem(x, set(y)) → =(x, y)
mem(x, union(y, z)) → or(mem(x, y), mem(x, z))

The set Q consists of the following terms:

or(true, x0)
or(x0, true)
or(false, false)
mem(x0, nil)
mem(x0, set(x1))
mem(x0, union(x1, x2))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MEM(x, union(y, z)) → MEM(x, z)
MEM(x, union(y, z)) → MEM(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MEM(x1, x2)  =  x2
union(x1, x2)  =  union(x1, x2)
or(x1, x2)  =  or
true  =  true
false  =  false
mem(x1, x2)  =  mem
nil  =  nil
set(x1)  =  set
=(x1, x2)  =  =

Recursive Path Order [RPO].
Precedence:

mem > or > true
mem > or > false
mem > =

The following usable rules [FROCOS05] were oriented:

or(true, y) → true
or(x, true) → true
or(false, false) → false
mem(x, nil) → false
mem(x, set(y)) → =(x, y)
mem(x, union(y, z)) → or(mem(x, y), mem(x, z))

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

or(true, y) → true
or(x, true) → true
or(false, false) → false
mem(x, nil) → false
mem(x, set(y)) → =(x, y)
mem(x, union(y, z)) → or(mem(x, y), mem(x, z))

The set Q consists of the following terms:

or(true, x0)
or(x0, true)
or(false, false)
mem(x0, nil)
mem(x0, set(x1))
mem(x0, union(x1, x2))

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

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE