Consider the TRS R consisting of the rewrite rules

1: le(0,Y) -> true
2: le(s(X),0) -> false
3: le(s(X),s(Y)) -> le(X,Y)
4: minus(0,Y) -> 0
5: minus(s(X),Y) -> ifMinus(le(s(X),Y),s(X),Y)
6: ifMinus(true,s(X),Y) -> 0
7: ifMinus(false,s(X),Y) -> s(minus(X,Y))
8: quot(0,s(Y)) -> 0
9: quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))

There are 6 dependency pairs:

10: LE(s(X),s(Y)) -> LE(X,Y)
11: MINUS(s(X),Y) -> IFMINUS(le(s(X),Y),s(X),Y)
12: MINUS(s(X),Y) -> LE(s(X),Y)
13: IFMINUS(false,s(X),Y) -> MINUS(X,Y)
14: QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
15: QUOT(s(X),s(Y)) -> MINUS(X,Y)

The approximated dependency graph contains 3 SCCs:
{10},
{11,13}
and {14}.

- Consider the SCC {10}.
There are no usable rules.
By taking the polynomial interpretation
[s](x) = x + 1
and [LE](x,y) = x + y + 1,
rule 10
is strictly decreasing.

- Consider the SCC {11,13}.
The usable rules are {1-3}.
By taking the polynomial interpretation
[0] = [false] = [true] = 1,
[s](x) = x + 1,
[le](x,y) = [MINUS](x,y) = x + y + 1
and [IFMINUS](x,y,z) = y + z + 1,
rule 11
is weakly decreasing and
the rules in {1-3,13}
are strictly decreasing.

- Consider the SCC {14}.
The usable rules are {1-7}.
By taking the polynomial interpretation
[0] = [false] = [true] = 1,
[minus](x,y) = x,
[s](x) = x + 1,
[le](x,y) = [QUOT](x,y) = x + y + 1
and [ifMinus](x,y,z) = y,
the rules in {4-7}
are weakly decreasing and
the rules in {1-3,14}
are strictly decreasing.

Hence the TRS is terminating.