YES Problem: p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3))) Proof: DP Processor: DPs: p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) TRS: p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3))) EDG Processor: DPs: p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) TRS: p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3))) graph: p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) -> p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) -> p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) -> p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) -> p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(b(x1),x3) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) -> p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) -> p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) SCC Processor: #sccs: 1 #rules: 2 #arcs: 6/9 DPs: p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(a(a(x0)),p(b(x1),x3)) p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) TRS: p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3))) Subterm Criterion Processor: simple projection: pi(p) = [1,1] pi(p#) = [1,1] problem: DPs: p#(a(x0),p(b(x1),p(a(x2),x3))) -> p#(x2,p(a(a(x0)),p(b(x1),x3))) TRS: p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3))) Semantic Labeling Processor: dimension: 2 usable rules: interpretation: [0] [b](x0) = [1], [1 1] [a](x0) = [0 0]x0, [1 1] [0 1] [1] [p](x0, x1) = [1 0]x0 + [0 1]x1 + [0] labeled: p# usable (for model): p# a p b argument filtering: pi(a) = 0 pi(b) = [] pi(p) = 0 pi(p#) = [] precedence: p# ~ p ~ b ~ a problem: DPs: TRS: p(a(x0),p(b(x1),p(a(x2),x3))) -> p(x2,p(a(a(x0)),p(b(x1),x3))) Qed