YES We show the termination of the TRS R: g(x,y) -> x g(x,y) -> y f(|0|(),|1|(),x) -> f(s(x),x,x) f(x,y,s(z)) -> s(f(|0|(),|1|(),z)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x) p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z) and R consists of: r1: g(x,y) -> x r2: g(x,y) -> y r3: f(|0|(),|1|(),x) -> f(s(x),x,x) r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z)) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x) p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z) and R consists of: r1: g(x,y) -> x r2: g(x,y) -> y r3: f(|0|(),|1|(),x) -> f(s(x),x,x) r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z)) The set of usable rules consists of r1, r2 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: f#_A(x1,x2,x3) = 2 |0|_A = 0 |1|_A = 1 s_A(x1) = max{0, x1 - 3} g_A(x1,x2) = max{x1 + 1, x2 + 1} base order 2: marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: f#_A(x1,x2,x3) = max{0, x1 - 13, x2 - 9, x3 - 9} |0|_A = 12 |1|_A = 8 s_A(x1) = max{7, x1 + 4} g_A(x1,x2) = max{x1, x2} f#!_A(x1,x2,x3) = max{0, x3 - 5} |0|!_A = 0 |1|!_A = 1 s!_A(x1) = max{0, x1 - 28} g!_A(x1,x2) = max{x1, x2} partial status: pi(f#) = [] pi(|0|) = [] pi(|1|) = [] pi(s) = [] pi(g) = [1, 2] The following rules are removed: p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x) and R consists of: r1: g(x,y) -> x r2: g(x,y) -> y r3: f(|0|(),|1|(),x) -> f(s(x),x,x) r4: f(x,y,s(z)) -> s(f(|0|(),|1|(),z)) The estimated dependency graph contains the following SCCs: (no SCCs)