YES Problem: t(o(x1)) -> m(a(x1)) t(e(x1)) -> n(s(x1)) a(l(x1)) -> a(t(x1)) o(m(a(x1))) -> t(e(n(x1))) s(a(x1)) -> l(a(t(o(m(a(t(e(x1)))))))) n(s(x1)) -> a(l(a(t(x1)))) Proof: String Reversal Processor: o(t(x1)) -> a(m(x1)) e(t(x1)) -> s(n(x1)) l(a(x1)) -> t(a(x1)) a(m(o(x1))) -> n(e(t(x1))) a(s(x1)) -> e(t(a(m(o(t(a(l(x1)))))))) s(n(x1)) -> t(a(l(a(x1)))) Bounds Processor: bound: 4 enrichment: match automaton: final states: {19,11,8,6,4,1} transitions: l1(58) -> 59* n2(62) -> 63* a4(116) -> 117* n0(10) -> 8* n0(2) -> 5* s0(5) -> 4* a3(108) -> 109* a3(103) -> 104* a3(94) -> 95* a3(106) -> 107* f80() -> 2* m0(15) -> 16* m0(2) -> 3* a2(76) -> 77* a2(71) -> 72* a2(89) -> 90* a2(91) -> 92* a2(74) -> 75* a0(12) -> 13* a0(20) -> 21* a0(2) -> 7* a0(3) -> 1* a0(16) -> 17* l2(90) -> 91* l2(75) -> 76* e0(18) -> 11* e0(9) -> 10* t4(117) -> 118* n1(31) -> 32* n1(34) -> 35* n1(48) -> 49* t0(2) -> 9* t0(21) -> 19* t0(13) -> 14* t0(17) -> 18* t0(7) -> 6* m1(22) -> 23* o0(14) -> 15* t3(109) -> 110* t3(104) -> 105* t3(95) -> 96* t2(72) -> 73* t2(92) -> 93* t2(77) -> 78* s1(35) -> 36* s1(32) -> 33* s2(63) -> 64* l0(7) -> 20* l0(2) -> 12* l3(107) -> 108* t1(44) -> 45* t1(46) -> 47* t1(60) -> 61* a1(23) -> 24* a1(43) -> 44* a1(59) -> 60* e1(47) -> 48* 43 -> 71* 24 -> 15* 74 -> 94* 17 -> 34* 46 -> 62* 64 -> 48* 93 -> 33,10 14 -> 46* 106 -> 116* 45 -> 20* 118 -> 108* 11 -> 104,90,7,44,72 2 -> 43,31 105 -> 91* 6 -> 12* 36 -> 11* 73 -> 59* 44 -> 58* 13 -> 22* 8 -> 104,90,7,44,72,1 49 -> 17* 31 -> 89* 62 -> 106* 34 -> 74* 110 -> 64,48 96 -> 76* 89 -> 103* 61 -> 4* 33 -> 10* 78 -> 36,11 problem: Qed