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, r2, r3, r4, r5, r6, r7, r8 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: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: app#_A(0,0) = 0 app#_A(0,1) = 0 app#_A(1,0) = 0 app#_A(1,1) = 0 fmap_1_A(0) = 1 fmap_1_A(1) = 1 fmap_2#_A(0,0) = 0 fmap_2#_A(0,1) = 0 fmap_2#_A(1,0) = 0 fmap_2#_A(1,1) = 0 fcons_2_A(0,0) = 1 fcons_2_A(0,1) = 1 fcons_2_A(1,0) = 1 fcons_2_A(1,1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 cons_0_A = 0 cons_1_A(0) = 1 cons_1_A(1) = 1 cons_2_A(0,0) = 0 cons_2_A(0,1) = 0 cons_2_A(1,0) = 0 cons_2_A(1,1) = 0 fcons_0_A = 1 fcons_1_A(0) = 0 fcons_1_A(1) = 0 fmap_0_A = 0 fmap_2_A(0,0) = 0 fmap_2_A(0,1) = 0 fmap_2_A(1,0) = 0 fmap_2_A(1,1) = 0 fnil_0_A = 0 nil_0_A = 0 app#!_A(0,0) = 0 app#!_A(0,1) = 0 app#!_A(1,0) = 1 app#!_A(1,1) = 0 fmap_1!_A(0) = 0 fmap_1!_A(1) = 0 fmap_2#!_A(0,0) = 1 fmap_2#!_A(0,1) = 0 fmap_2#!_A(1,0) = 1 fmap_2#!_A(1,1) = 0 fcons_2!_A(0,0) = 0 fcons_2!_A(0,1) = 0 fcons_2!_A(1,0) = 1 fcons_2!_A(1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 cons_0!_A = 1 cons_1!_A(0) = 1 cons_1!_A(1) = 0 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 1 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 fcons_0!_A = 0 fcons_1!_A(0) = 0 fcons_1!_A(1) = 0 fmap_0!_A = 0 fmap_2!_A(0,0) = 0 fmap_2!_A(0,1) = 0 fmap_2!_A(1,0) = 0 fmap_2!_A(1,1) = 0 fnil_0!_A = 1 nil_0!_A = 0 partial status: pi(app#) = [1] pi(fmap_1) = [1] pi(fmap_2#) = [1] pi(fcons_2) = [1, 2] pi(app) = [1, 2] pi(cons_0) = [] pi(cons_1) = [] pi(cons_2) = [] pi(fcons_0) = [] pi(fcons_1) = [1] pi(fmap_0) = [] pi(fmap_2) = [] pi(fnil_0) = [] pi(nil_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, r2, r3, r4, r5, r6, r7, r8 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: model-based reduction triple (=A, >=A!, >A!): finite algebra: carrier: {0,1} interpretations: fmap_2#_A(0,0) = 0 fmap_2#_A(0,1) = 0 fmap_2#_A(1,0) = 0 fmap_2#_A(1,1) = 0 fcons_2_A(0,0) = 1 fcons_2_A(0,1) = 1 fcons_2_A(1,0) = 1 fcons_2_A(1,1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 cons_0_A = 0 cons_1_A(0) = 1 cons_1_A(1) = 1 cons_2_A(0,0) = 0 cons_2_A(0,1) = 0 cons_2_A(1,0) = 0 cons_2_A(1,1) = 0 fcons_0_A = 1 fcons_1_A(0) = 0 fcons_1_A(1) = 0 fmap_0_A = 0 fmap_1_A(0) = 1 fmap_1_A(1) = 1 fmap_2_A(0,0) = 0 fmap_2_A(0,1) = 0 fmap_2_A(1,0) = 0 fmap_2_A(1,1) = 0 fnil_0_A = 0 nil_0_A = 0 fmap_2#!_A(0,0) = 0 fmap_2#!_A(0,1) = 0 fmap_2#!_A(1,0) = 0 fmap_2#!_A(1,1) = 0 fcons_2!_A(0,0) = 1 fcons_2!_A(0,1) = 1 fcons_2!_A(1,0) = 1 fcons_2!_A(1,1) = 1 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 0 cons_0!_A = 0 cons_1!_A(0) = 1 cons_1!_A(1) = 1 cons_2!_A(0,0) = 0 cons_2!_A(0,1) = 0 cons_2!_A(1,0) = 0 cons_2!_A(1,1) = 0 fcons_0!_A = 1 fcons_1!_A(0) = 0 fcons_1!_A(1) = 0 fmap_0!_A = 1 fmap_1!_A(0) = 1 fmap_1!_A(1) = 1 fmap_2!_A(0,0) = 0 fmap_2!_A(0,1) = 0 fmap_2!_A(1,0) = 0 fmap_2!_A(1,1) = 0 fnil_0!_A = 1 nil_0!_A = 0 partial status: pi(fmap_2#) = [1] pi(fcons_2) = [2] pi(app) = [2] pi(cons_0) = [] pi(cons_1) = [1] pi(cons_2) = [] pi(fcons_0) = [] pi(fcons_1) = [] pi(fmap_0) = [] pi(fmap_1) = [1] pi(fmap_2) = [] pi(fnil_0) = [] pi(nil_0) = [] The following rules are removed: p1 Then no dependency pair remains.