YES Problem: times(x,plus(y,1())) -> plus(times(x,plus(y,times(1(),0()))),x) times(x,1()) -> x plus(x,0()) -> x times(x,0()) -> 0() Proof: DP Processor: DPs: times#(x,plus(y,1())) -> times#(1(),0()) times#(x,plus(y,1())) -> plus#(y,times(1(),0())) times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) times#(x,plus(y,1())) -> plus#(times(x,plus(y,times(1(),0()))),x) TRS: times(x,plus(y,1())) -> plus(times(x,plus(y,times(1(),0()))),x) times(x,1()) -> x plus(x,0()) -> x times(x,0()) -> 0() EDG Processor: DPs: times#(x,plus(y,1())) -> times#(1(),0()) times#(x,plus(y,1())) -> plus#(y,times(1(),0())) times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) times#(x,plus(y,1())) -> plus#(times(x,plus(y,times(1(),0()))),x) TRS: times(x,plus(y,1())) -> plus(times(x,plus(y,times(1(),0()))),x) times(x,1()) -> x plus(x,0()) -> x times(x,0()) -> 0() graph: times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) -> times#(x,plus(y,1())) -> times#(1(),0()) times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) -> times#(x,plus(y,1())) -> plus#(y,times(1(),0())) times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) -> times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) -> times#(x,plus(y,1())) -> plus#(times(x,plus(y,times(1(),0()))),x) SCC Processor: #sccs: 1 #rules: 1 #arcs: 4/16 DPs: times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) TRS: times(x,plus(y,1())) -> plus(times(x,plus(y,times(1(),0()))),x) times(x,1()) -> x plus(x,0()) -> x times(x,0()) -> 0() Usable Rule Processor: DPs: times#(x,plus(y,1())) -> times#(x,plus(y,times(1(),0()))) TRS: times(x,0()) -> 0() plus(x,0()) -> x Bounds Processor: bound: 1 enrichment: match-dp automaton: final states: {6} transitions: times{#,0}(5,8) -> 6* plus0(3,1) -> 4* plus0(3,3) -> 4* plus0(4,2) -> 4* plus0(4,4) -> 4* plus0(5,7) -> 8* plus0(1,2) -> 4* plus0(1,4) -> 4* plus0(2,1) -> 4* plus0(2,3) -> 4* plus0(3,2) -> 4* plus0(3,4) -> 4* plus0(4,1) -> 4* plus0(4,3) -> 4* plus0(1,1) -> 4* plus0(1,3) -> 4* plus0(2,2) -> 4* plus0(2,4) -> 4* 10() -> 3* times0(3,1) -> 7,2 times0(3,3) -> 2* times0(4,2) -> 2* times0(4,4) -> 2* times0(1,2) -> 2* times0(1,4) -> 2* times0(2,1) -> 2* times0(2,3) -> 2* times0(3,2) -> 2* times0(3,4) -> 2* times0(4,1) -> 2* times0(4,3) -> 2* times0(1,1) -> 2* times0(1,3) -> 2* times0(2,2) -> 2* times0(2,4) -> 2* 00() -> 7,2,1 times{#,1}(5,16) -> 6* plus1(3,15) -> 16* plus1(4,15) -> 16* times1(14,13) -> 15* 11() -> 14* 01() -> 15,13 1 -> 4,5 2 -> 4,5 3 -> 4,5 4 -> 16,5 5 -> 8* problem: DPs: TRS: times(x,0()) -> 0() plus(x,0()) -> x Qed