YES We show the termination of the TRS R: app(app(fmap(),fnil()),x) -> nil() app(app(fmap(),app(app(fcons(),f),t)),x) -> app(app(cons(),app(f,x)),app(app(fmap(),t),x)) Instead we prove termination of the uncurried system: app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(fcons_0(),y) -> fcons_1(y) app(fcons_1(x1),y) -> fcons_2(x1,y) app(fmap_0(),y) -> fmap_1(y) app(fmap_1(x1),y) -> fmap_2(x1,y) fmap_2(fnil_0(),x) -> nil_0() fmap_2(fcons_2(f,t),x) -> cons_2(app(f,x),fmap_2(t,x)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(fmap_1(x1),y) -> fmap_2#(x1,y) p2: fmap_2#(fcons_2(f,t),x) -> app#(f,x) p3: fmap_2#(fcons_2(f,t),x) -> fmap_2#(t,x) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(fcons_0(),y) -> fcons_1(y) r4: app(fcons_1(x1),y) -> fcons_2(x1,y) r5: app(fmap_0(),y) -> fmap_1(y) r6: app(fmap_1(x1),y) -> fmap_2(x1,y) r7: fmap_2(fnil_0(),x) -> nil_0() r8: fmap_2(fcons_2(f,t),x) -> cons_2(app(f,x),fmap_2(t,x)) The estimated dependency graph contains the following SCCs: {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(fmap_1(x1),y) -> fmap_2#(x1,y) p2: fmap_2#(fcons_2(f,t),x) -> fmap_2#(t,x) p3: fmap_2#(fcons_2(f,t),x) -> app#(f,x) and R consists of: r1: app(cons_0(),y) -> cons_1(y) r2: app(cons_1(x1),y) -> cons_2(x1,y) r3: app(fcons_0(),y) -> fcons_1(y) r4: app(fcons_1(x1),y) -> fcons_2(x1,y) r5: app(fmap_0(),y) -> fmap_1(y) r6: app(fmap_1(x1),y) -> fmap_2(x1,y) r7: fmap_2(fnil_0(),x) -> nil_0() r8: fmap_2(fcons_2(f,t),x) -> cons_2(app(f,x),fmap_2(t,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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 fmap_1_A(x1) = x1 + 3 fmap_2#_A(x1,x2) = x1 fcons_2_A(x1,x2) = x1 + x2 + 5 app#!_A(x1,x2) = 4 fmap_1!_A(x1) = 3 fmap_2#!_A(x1,x2) = 4 fcons_2!_A(x1,x2) = 5 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 15, x2 - 1} fmap_1_A(x1) = x1 + 1 fmap_2#_A(x1,x2) = max{0, x1 - 14, x2 - 1} fcons_2_A(x1,x2) = max{x1 + 11, x2 + 13} app#!_A(x1,x2) = max{10, x1 - 2} fmap_1!_A(x1) = max{5, x1 + 3} fmap_2#!_A(x1,x2) = max{6, x1 - 2} fcons_2!_A(x1,x2) = max{4, x2 + 2} partial status: pi(app#) = [2] pi(fmap_1) = [] pi(fmap_2#) = [2] pi(fcons_2) = [1, 2] The following rules are removed: p1, p2, p3 Then no dependency pair remains.