Problem: a__f(0()) -> cons(0(),f(s(0()))) a__f(s(0())) -> a__f(a__p(s(0()))) a__p(s(0())) -> 0() mark(f(X)) -> a__f(mark(X)) mark(p(X)) -> a__p(mark(X)) mark(0()) -> 0() mark(cons(X1,X2)) -> cons(mark(X1),X2) mark(s(X)) -> s(mark(X)) a__f(X) -> f(X) a__p(X) -> p(X) Proof: Bounds Processor: bound: 3 enrichment: match automaton: final states: {8,7,6} transitions: f3(27) -> 28* f3(24) -> 18,8 p1(5) -> 7* p1(2) -> 7* p1(4) -> 7* p1(1) -> 7* p1(3) -> 7* 03() -> 24* f1(10) -> 11* f1(5) -> 6* f1(2) -> 6* f1(4) -> 6* f1(1) -> 6* f1(3) -> 6* cons3(24,28) -> 18,8 s1(9) -> 10* s1(18) -> 18,8 s3(24) -> 27* mark1(5) -> 18* mark1(2) -> 18* mark1(4) -> 18* mark1(1) -> 18* mark1(3) -> 18* cons1(18,1) -> 18,8 cons1(18,3) -> 18,8 cons1(18,5) -> 18,8 cons1(18,2) -> 18,8 cons1(18,4) -> 18,8 cons1(9,11) -> 6* 01() -> 18,8,7,9 a__p1(10) -> 16* a__p1(18) -> 18,8 a__f1(16) -> 6* a__f1(18) -> 18,8 p2(10) -> 16* p2(18) -> 18,8 a__f0(5) -> 6* a__f0(2) -> 6* a__f0(4) -> 6* a__f0(1) -> 6* a__f0(3) -> 6* f2(19) -> 20* f2(16) -> 6* f2(18) -> 18,8 00() -> 1* 02() -> 8,18,16 cons0(3,1) -> 2* cons0(3,3) -> 2* cons0(3,5) -> 2* cons0(4,2) -> 2* cons0(4,4) -> 2* cons0(5,1) -> 2* cons0(5,3) -> 2* cons0(5,5) -> 2* cons0(1,2) -> 2* cons0(1,4) -> 2* cons0(2,1) -> 2* cons0(2,3) -> 2* cons0(2,5) -> 2* cons0(3,2) -> 2* cons0(3,4) -> 2* cons0(4,1) -> 2* cons0(4,3) -> 2* cons0(4,5) -> 2* cons0(5,2) -> 2* cons0(5,4) -> 2* cons0(1,1) -> 2* cons0(1,3) -> 2* cons0(1,5) -> 2* cons0(2,2) -> 2* cons0(2,4) -> 2* a__f2(24) -> 8,18 f0(5) -> 3* f0(2) -> 3* f0(4) -> 3* f0(1) -> 3* f0(3) -> 3* a__p2(19) -> 24* s0(5) -> 4* s0(2) -> 4* s0(4) -> 4* s0(1) -> 4* s0(3) -> 4* s2(16) -> 19* a__p0(5) -> 7* a__p0(2) -> 7* a__p0(4) -> 7* a__p0(1) -> 7* a__p0(3) -> 7* cons2(16,20) -> 6,18,8 mark0(5) -> 8* mark0(2) -> 8* mark0(4) -> 8* mark0(1) -> 8* mark0(3) -> 8* p3(19) -> 24* p0(5) -> 5* p0(2) -> 5* p0(4) -> 5* p0(1) -> 5* p0(3) -> 5* problem: Qed