YES(?,O(n^1)) Problem: h(x,c(y,z)) -> h(c(s(y),x),z) h(c(s(x),c(s(0()),y)),z) -> h(y,c(s(0()),c(x,z))) Proof: Bounds Processor: bound: 2 enrichment: match automaton: final states: {4} transitions: h1(12,16) -> 4* h1(2,16) -> 4* h1(8,1) -> 4* h1(8,3) -> 4* h1(1,16) -> 4* h1(12,1) -> 4* h1(12,3) -> 4* h1(8,2) -> 4* h1(8,16) -> 4* h1(3,16) -> 4* h1(12,2) -> 4* c1(2,16) -> 13* c1(3,1) -> 13* c1(3,3) -> 13* c1(7,28) -> 8* c1(7,30) -> 8* c1(7,34) -> 8* c1(15,13) -> 16* c1(11,2) -> 8* c1(1,2) -> 13* c1(11,8) -> 8* c1(11,12) -> 8* c1(1,16) -> 13* c1(7,1) -> 8* c1(2,1) -> 13* c1(7,3) -> 8* c1(2,3) -> 13* c1(11,28) -> 8* c1(11,30) -> 8* c1(11,34) -> 8* c1(3,2) -> 13* c1(3,16) -> 13* c1(15,16) -> 16* c1(11,1) -> 12* c1(1,1) -> 13* c1(11,3) -> 8* c1(1,3) -> 13* c1(7,2) -> 8* c1(2,2) -> 13* c1(7,8) -> 8* c1(7,12) -> 8* s1(2) -> 7* s1(14) -> 15* s1(1) -> 11* s1(3) -> 7* 01() -> 14* h2(22,16) -> 4* h2(28,1) -> 4* h2(28,3) -> 4* h2(34,2) -> 4* h2(34,16) -> 4* h2(30,1) -> 4* h2(30,3) -> 4* h2(22,13) -> 4* h2(28,2) -> 4* h2(28,16) -> 4* h2(34,1) -> 4* h2(34,3) -> 4* h2(30,2) -> 4* h2(30,16) -> 4* h0(3,1) -> 4* h0(3,3) -> 4* h0(1,2) -> 4* h0(2,1) -> 4* h0(2,3) -> 4* h0(3,2) -> 4* h0(1,1) -> 4* h0(1,3) -> 4* h0(2,2) -> 4* c2(27,22) -> 28* c2(29,22) -> 30* c2(21,2) -> 22* c2(21,8) -> 22* c2(21,12) -> 22* c2(21,22) -> 22* c2(21,28) -> 22* c2(21,30) -> 22* c2(21,34) -> 22* c2(33,22) -> 34* c2(21,1) -> 22* c2(21,3) -> 22* c0(3,1) -> 1* c0(3,3) -> 1* c0(1,2) -> 1* c0(2,1) -> 1* c0(2,3) -> 1* c0(3,2) -> 1* c0(1,1) -> 1* c0(1,3) -> 1* c0(2,2) -> 1* s2(15) -> 21* s2(2) -> 27* s2(1) -> 33* s2(3) -> 29* s0(2) -> 2* s0(1) -> 2* s0(3) -> 2* 00() -> 3* problem: Qed