YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
minus(minus(x:S)) -> x:S
minus(f(x:S,y:S)) -> f(minus(y:S),minus(x:S))
minus(h(x:S)) -> h(minus(x:S))
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 MINUS(f(x:S,y:S)) -> MINUS(x:S)
 MINUS(f(x:S,y:S)) -> MINUS(y:S)
 MINUS(h(x:S)) -> MINUS(x:S)
-> Rules:
 minus(minus(x:S)) -> x:S
 minus(f(x:S,y:S)) -> f(minus(y:S),minus(x:S))
 minus(h(x:S)) -> h(minus(x:S))

Problem 1: 

SCC Processor:
-> Pairs:
 MINUS(f(x:S,y:S)) -> MINUS(x:S)
 MINUS(f(x:S,y:S)) -> MINUS(y:S)
 MINUS(h(x:S)) -> MINUS(x:S)
-> Rules:
 minus(minus(x:S)) -> x:S
 minus(f(x:S,y:S)) -> f(minus(y:S),minus(x:S))
 minus(h(x:S)) -> h(minus(x:S))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MINUS(f(x:S,y:S)) -> MINUS(x:S)
 MINUS(f(x:S,y:S)) -> MINUS(y:S)
 MINUS(h(x:S)) -> MINUS(x:S)
->->-> Rules:
 minus(minus(x:S)) -> x:S
 minus(f(x:S,y:S)) -> f(minus(y:S),minus(x:S))
 minus(h(x:S)) -> h(minus(x:S))

Problem 1: 

Subterm Processor:
-> Pairs:
 MINUS(f(x:S,y:S)) -> MINUS(x:S)
 MINUS(f(x:S,y:S)) -> MINUS(y:S)
 MINUS(h(x:S)) -> MINUS(x:S)
-> Rules:
 minus(minus(x:S)) -> x:S
 minus(f(x:S,y:S)) -> f(minus(y:S),minus(x:S))
 minus(h(x:S)) -> h(minus(x:S))
->Projection:
 pi(MINUS) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 minus(minus(x:S)) -> x:S
 minus(f(x:S,y:S)) -> f(minus(y:S),minus(x:S))
 minus(h(x:S)) -> h(minus(x:S))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.