YES We show the termination of the TRS R: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p4: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) p5: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y) p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x) p7: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a()))))) p8: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p9: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p10: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a()))) p11: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) p12: c#(f(c(a(),y,a())),x,z) -> b#(x,a()) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p5: c#(f(c(a(),y,a())),x,z) -> f#(b(y,b(x,a()))) p6: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x) p7: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y) p8: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p9: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p10: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a()))))) p11: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The set of usable rules consists of r1, r2, r3 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: b#_A(x1,x2) = (4,8) a_A() = (4,7) b_A(x1,x2) = (5,16) c_A(x1,x2,x3) = ((0,0),(1,0)) x1 + (8,6) c#_A(x1,x2,x3) = (4,8) f_A(x1) = ((1,0),(0,0)) x1 + (1,0) f#_A(x1) = ((0,0),(1,0)) x1 + (4,1) precedence: a = b > c > f > b# = c# > f# partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [] pi(c#) = [] pi(f) = [] pi(f#) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: b#_A(x1,x2) = (1,3) a_A() = (2,3) b_A(x1,x2) = (1,3) c_A(x1,x2,x3) = (0,2) c#_A(x1,x2,x3) = (1,3) f_A(x1) = (0,1) f#_A(x1) = (1,3) precedence: b# = a = b = c = c# = f# > f partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [] pi(c#) = [] pi(f) = [] pi(f#) = [] The next rules are strictly ordered: p5 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p5: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x) p6: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y) p7: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p8: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p9: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a()))))) p10: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p4: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a()))))) p5: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) p6: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y) p7: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p8: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p9: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) p10: f#(c(a(),b(b(z,a()),y),x)) -> b#(z,x) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The set of usable rules consists of r1, r2, r3 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: b#_A(x1,x2) = (3,2) a_A() = (6,12) b_A(x1,x2) = ((0,0),(1,0)) x1 + (1,5) c_A(x1,x2,x3) = (5,4) c#_A(x1,x2,x3) = (3,2) f_A(x1) = ((1,0),(0,0)) x1 + (3,3) f#_A(x1) = ((1,0),(0,0)) x1 + (2,1) precedence: f# > b = c > a > f > b# = c# partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [] pi(c#) = [] pi(f) = [] pi(f#) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: b#_A(x1,x2) = (0,0) a_A() = (1,4) b_A(x1,x2) = (3,3) c_A(x1,x2,x3) = (2,2) c#_A(x1,x2,x3) = (0,0) f_A(x1) = (0,1) f#_A(x1) = (0,0) precedence: b = c > b# = a = c# = f = f# partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [] pi(c#) = [] pi(f) = [] pi(f#) = [] The next rules are strictly ordered: p10 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p4: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a()))))) p5: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) p6: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y) p7: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p8: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p9: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p5: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p6: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p7: c#(f(c(a(),y,a())),x,z) -> f#(b(b(z,z),f(b(y,b(x,a()))))) p8: f#(c(a(),b(b(z,a()),y),x)) -> c#(x,b(z,x),y) p9: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The set of usable rules consists of r1, r2, r3 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: b#_A(x1,x2) = (7,7) a_A() = (9,3) b_A(x1,x2) = ((0,0),(1,0)) x1 + (1,0) c_A(x1,x2,x3) = (6,2) c#_A(x1,x2,x3) = (7,7) f_A(x1) = x1 + (4,4) f#_A(x1) = ((1,0),(0,0)) x1 + (2,4) precedence: b# = a = b = c = c# = f = f# partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [] pi(c#) = [] pi(f) = [] pi(f#) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: b#_A(x1,x2) = (1,3) a_A() = (0,0) b_A(x1,x2) = (0,0) c_A(x1,x2,x3) = (0,2) c#_A(x1,x2,x3) = (1,3) f_A(x1) = (0,1) f#_A(x1) = (2,4) precedence: b = c > b# = c# > a = f = f# partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [] pi(c#) = [] pi(f) = [] pi(f#) = [] The next rules are strictly ordered: p7, p8 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p3: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) p4: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p5: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p6: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p7: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6} {p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: b#(a(),b(c(z,x,y),a())) -> b#(b(z,c(y,z,a())),x) p2: b#(a(),b(c(z,x,y),a())) -> b#(z,c(y,z,a())) p3: b#(a(),b(c(z,x,y),a())) -> c#(y,z,a()) p4: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p5: c#(f(c(a(),y,a())),x,z) -> b#(z,z) p6: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The set of usable rules consists of r1, r2, r3 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: b#_A(x1,x2) = x2 + (1,0) a_A() = (10,4) b_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,0)) x2 + (0,4) c_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((0,0),(1,1)) x2 + ((1,1),(1,1)) x3 + (21,29) c#_A(x1,x2,x3) = ((1,1),(0,1)) x1 + ((1,1),(0,1)) x2 + ((1,0),(1,1)) x3 + (10,14) f_A(x1) = (2,4) precedence: a = c = f > b > c# > b# partial status: pi(b#) = [2] pi(a) = [] pi(b) = [1] pi(c) = [1, 3] pi(c#) = [1, 2, 3] pi(f) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: b#_A(x1,x2) = ((0,1),(0,1)) x2 + (13,0) a_A() = (1,1) b_A(x1,x2) = x1 + (7,0) c_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,0)) x3 + (8,0) c#_A(x1,x2,x3) = ((1,0),(0,0)) x1 + ((0,1),(0,1)) x2 + (7,0) f_A(x1) = (7,0) precedence: a > b# > c = f > b = c# partial status: pi(b#) = [] pi(a) = [] pi(b) = [] pi(c) = [1] pi(c#) = [] pi(f) = [] The next rules are strictly ordered: p1, p2, p3, p5 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: c#(f(c(a(),y,a())),x,z) -> b#(b(z,z),f(b(y,b(x,a())))) p2: c#(f(c(a(),y,a())),x,z) -> b#(y,b(x,a())) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The estimated dependency graph contains the following SCCs: (no SCCs) -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f#(c(a(),b(b(z,a()),y),x)) -> f#(c(x,b(z,x),y)) and R consists of: r1: b(a(),b(c(z,x,y),a())) -> b(b(z,c(y,z,a())),x) r2: f(c(a(),b(b(z,a()),y),x)) -> f(c(x,b(z,x),y)) r3: c(f(c(a(),y,a())),x,z) -> f(b(b(z,z),f(b(y,b(x,a()))))) The set of usable rules consists of r1, r2, r3 Take the reduction pair: lexicographic combination of reduction pairs: 1. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: f#_A(x1) = ((1,0),(0,0)) x1 + (1,8) c_A(x1,x2,x3) = ((0,0),(1,1)) x1 + ((1,0),(1,1)) x2 + ((0,1),(0,0)) x3 + (9,12) a_A() = (3,0) b_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(1,0)) x2 + (5,2) f_A(x1) = (4,1) precedence: f# = c = a = b = f partial status: pi(f#) = [] pi(c) = [] pi(a) = [] pi(b) = [] pi(f) = [] 2. weighted path order base order: matrix interpretations: carrier: N^2 order: standard order interpretations: f#_A(x1) = (0,0) c_A(x1,x2,x3) = x2 + (1,2) a_A() = (3,3) b_A(x1,x2) = (1,1) f_A(x1) = (0,0) precedence: a > c = b > f# = f partial status: pi(f#) = [] pi(c) = [2] pi(a) = [] pi(b) = [] pi(f) = [] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.