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 r1, r3, r5 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 fmap_1_A(x1) = x1 + 2 fmap_2#_A(x1,x2) = x1 fcons_2_A(x1,x2) = x1 + x2 + 1 app_A(x1,x2) = x2 + 5 cons_0_A() = 2 cons_1_A(x1) = 1 fcons_0_A() = 3 fcons_1_A(x1) = 2 fmap_0_A() = 1 base order 2: precedence: fcons_0 > fcons_1 > fcons_2 = app = cons_0 > fmap_1 = cons_1 > fmap_2# = fmap_0 > app# partial status: pi(app#) = [1] pi(fmap_1) = [1] pi(fmap_2#) = [] pi(fcons_2) = [] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [] pi(fcons_0) = [] pi(fcons_1) = [] pi(fmap_0) = [] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: fmap_2#(fcons_2(f,t),x) -> fmap_2#(t,x) p2: 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 estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: 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 set of usable rules consists of r1, r3, r5 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: fmap_2#_A(x1,x2) = x1 fcons_2_A(x1,x2) = x1 + x2 + 6 app_A(x1,x2) = 4 cons_0_A() = 2 cons_1_A(x1) = 1 fcons_0_A() = 2 fcons_1_A(x1) = 1 fmap_0_A() = 1 fmap_1_A(x1) = 2 base order 2: precedence: cons_0 = fcons_0 = fmap_0 > fmap_1 > fcons_2 > fmap_2# = app > cons_1 = fcons_1 partial status: pi(fmap_2#) = [] pi(fcons_2) = [2] pi(app) = [] pi(cons_0) = [] pi(cons_1) = [] pi(fcons_0) = [] pi(fcons_1) = [] pi(fmap_0) = [] pi(fmap_1) = [] The following rules are removed: p1 Then no dependency pair remains.