YES Problem: a(b(c(x1))) -> c(b(a(x1))) C(B(A(x1))) -> A(B(C(x1))) b(a(C(x1))) -> C(a(b(x1))) c(A(B(x1))) -> B(A(c(x1))) A(c(b(x1))) -> b(c(A(x1))) B(C(a(x1))) -> a(C(B(x1))) a(A(x1)) -> x1 A(a(x1)) -> x1 b(B(x1)) -> x1 B(b(x1)) -> x1 c(C(x1)) -> x1 C(c(x1)) -> x1 Proof: Matrix Interpretation Processor: dim=1 interpretation: [B](x0) = x0 + 14, [b](x0) = x0 + 3, [A](x0) = x0 + 3, [C](x0) = x0 + 5, [c](x0) = x0 + 12, [a](x0) = x0 orientation: a(b(c(x1))) = x1 + 15 >= x1 + 15 = c(b(a(x1))) C(B(A(x1))) = x1 + 22 >= x1 + 22 = A(B(C(x1))) b(a(C(x1))) = x1 + 8 >= x1 + 8 = C(a(b(x1))) c(A(B(x1))) = x1 + 29 >= x1 + 29 = B(A(c(x1))) A(c(b(x1))) = x1 + 18 >= x1 + 18 = b(c(A(x1))) B(C(a(x1))) = x1 + 19 >= x1 + 19 = a(C(B(x1))) a(A(x1)) = x1 + 3 >= x1 = x1 A(a(x1)) = x1 + 3 >= x1 = x1 b(B(x1)) = x1 + 17 >= x1 = x1 B(b(x1)) = x1 + 17 >= x1 = x1 c(C(x1)) = x1 + 17 >= x1 = x1 C(c(x1)) = x1 + 17 >= x1 = x1 problem: a(b(c(x1))) -> c(b(a(x1))) C(B(A(x1))) -> A(B(C(x1))) b(a(C(x1))) -> C(a(b(x1))) c(A(B(x1))) -> B(A(c(x1))) A(c(b(x1))) -> b(c(A(x1))) B(C(a(x1))) -> a(C(B(x1))) String Reversal Processor: c(b(a(x1))) -> a(b(c(x1))) A(B(C(x1))) -> C(B(A(x1))) C(a(b(x1))) -> b(a(C(x1))) B(A(c(x1))) -> c(A(B(x1))) b(c(A(x1))) -> A(c(b(x1))) a(C(B(x1))) -> B(C(a(x1))) Bounds Processor: bound: 0 enrichment: match automaton: final states: {17,14,11,8,5,1} transitions: B0(2) -> 12* B0(6) -> 7* B0(19) -> 17* a0(4) -> 1* a0(9) -> 10* a0(2) -> 18* C0(7) -> 5* C0(18) -> 19* C0(2) -> 9* c0(2) -> 3* c0(13) -> 11* c0(15) -> 16* b0(2) -> 15* b0(10) -> 8* b0(3) -> 4* A0(2) -> 6* A0(12) -> 13* A0(16) -> 14* f60() -> 2* 17 -> 18,10 14 -> 15,4 11 -> 12,7 8 -> 9,19 1 -> 3,16 5 -> 6,13 problem: Qed