YES

Problem 1: 

(VAR v_NonEmpty:S X:S X1:S X2:S X3:S)
(RULES
a__c -> b
a__c -> c
a__f(b,X:S,c) -> a__f(X:S,a__c,X:S)
a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
mark(b) -> b
mark(c) -> a__c
mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 A__F(b,X:S,c) -> A__C
 A__F(b,X:S,c) -> A__F(X:S,a__c,X:S)
 MARK(c) -> A__C
 MARK(f(X1:S,X2:S,X3:S)) -> A__F(X1:S,mark(X2:S),X3:S)
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
-> Rules:
 a__c -> b
 a__c -> c
 a__f(b,X:S,c) -> a__f(X:S,a__c,X:S)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(b) -> b
 mark(c) -> a__c
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)

Problem 1: 

SCC Processor:
-> Pairs:
 A__F(b,X:S,c) -> A__C
 A__F(b,X:S,c) -> A__F(X:S,a__c,X:S)
 MARK(c) -> A__C
 MARK(f(X1:S,X2:S,X3:S)) -> A__F(X1:S,mark(X2:S),X3:S)
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
-> Rules:
 a__c -> b
 a__c -> c
 a__f(b,X:S,c) -> a__f(X:S,a__c,X:S)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(b) -> b
 mark(c) -> a__c
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
->->-> Rules:
 a__c -> b
 a__c -> c
 a__f(b,X:S,c) -> a__f(X:S,a__c,X:S)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(b) -> b
 mark(c) -> a__c
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)

Problem 1: 

Subterm Processor:
-> Pairs:
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
-> Rules:
 a__c -> b
 a__c -> c
 a__f(b,X:S,c) -> a__f(X:S,a__c,X:S)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(b) -> b
 mark(c) -> a__c
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Projection:
 pi(MARK) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 a__c -> b
 a__c -> c
 a__f(b,X:S,c) -> a__f(X:S,a__c,X:S)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(b) -> b
 mark(c) -> a__c
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.