YES We show the termination of the TRS R: app(app(F(),app(app(F(),f),x)),x) -> app(app(F(),app(G(),app(app(F(),f),x))),app(f,x)) Instead we prove termination of the uncurried system: app(F_0(),y) -> F_1(y) app(F_1(x1),y) -> F_2(x1,y) app(G_0(),y) -> G_1(y) F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(F_1(x1),y) -> F_2#(x1,y) p2: F_2#(F_2(f,x),x) -> F_2#(G_1(F_2(f,x)),app(f,x)) p3: F_2#(F_2(f,x),x) -> app#(f,x) and R consists of: r1: app(F_0(),y) -> F_1(y) r2: app(F_1(x1),y) -> F_2(x1,y) r3: app(G_0(),y) -> G_1(y) r4: F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x)) The estimated dependency graph contains the following SCCs: {p1, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(F_1(x1),y) -> F_2#(x1,y) p2: F_2#(F_2(f,x),x) -> app#(f,x) and R consists of: r1: app(F_0(),y) -> F_1(y) r2: app(F_1(x1),y) -> F_2(x1,y) r3: app(G_0(),y) -> G_1(y) r4: F_2(F_2(f,x),x) -> F_2(G_1(F_2(f,x)),app(f,x)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 39, x2 - 42} F_1_A(x1) = x1 + 38 F_2#_A(x1,x2) = max{0, x1 - 12, x2 - 42} F_2_A(x1,x2) = max{11, x1 - 22, x2 - 30} app#!_A(x1,x2) = max{1, x1 - 35, x2 - 43} F_1!_A(x1) = x1 + 5 F_2#!_A(x1,x2) = max{2, x1 - 12, x2 - 44} F_2!_A(x1,x2) = max{x1 + 6, x2 + 5} 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 + x2 F_1_A(x1) = x1 + 3 F_2#_A(x1,x2) = x1 F_2_A(x1,x2) = x1 + x2 + 5 app#!_A(x1,x2) = 1 F_1!_A(x1) = 3 F_2#!_A(x1,x2) = x1 F_2!_A(x1,x2) = x1 + x2 + 2 partial status: pi(app#) = [] pi(F_1) = [] pi(F_2#) = [] pi(F_2) = [2] The following rules are removed: p1, p2 Then no dependency pair remains.