YES We show the termination of the TRS R: app(app(mapbt(),f),app(leaf(),x)) -> app(leaf(),app(f,x)) app(app(mapbt(),f),app(app(app(branch(),x),l),r)) -> app(app(app(branch(),app(f,x)),app(app(mapbt(),f),l)),app(app(mapbt(),f),r)) Instead we prove termination of the uncurried system: app(branch_0(),y) -> branch_1(y) app(branch_1(x1),y) -> branch_2(x1,y) app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) app(leaf_0(),y) -> leaf_1(y) app(mapbt_0(),y) -> mapbt_1(y) app(mapbt_1(x1),y) -> mapbt_2(x1,y) mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(mapbt_1(x1),y) -> mapbt_2#(x1,y) p2: mapbt_2#(f,leaf_1(x)) -> app#(f,x) p3: mapbt_2#(f,branch_3(x,l,r)) -> app#(f,x) p4: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l) p5: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r) and R consists of: r1: app(branch_0(),y) -> branch_1(y) r2: app(branch_1(x1),y) -> branch_2(x1,y) r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) r4: app(leaf_0(),y) -> leaf_1(y) r5: app(mapbt_0(),y) -> mapbt_1(y) r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y) r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(mapbt_1(x1),y) -> mapbt_2#(x1,y) p2: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r) p3: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l) p4: mapbt_2#(f,branch_3(x,l,r)) -> app#(f,x) p5: mapbt_2#(f,leaf_1(x)) -> app#(f,x) and R consists of: r1: app(branch_0(),y) -> branch_1(y) r2: app(branch_1(x1),y) -> branch_2(x1,y) r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) r4: app(leaf_0(),y) -> leaf_1(y) r5: app(mapbt_0(),y) -> mapbt_1(y) r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y) r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) 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 mapbt_1_A(0) = 0 mapbt_1_A(1) = 0 mapbt_2#_A(0,0) = 0 mapbt_2#_A(0,1) = 0 mapbt_2#_A(1,0) = 0 mapbt_2#_A(1,1) = 0 branch_3_A(0,0,0) = 1 branch_3_A(0,0,1) = 1 branch_3_A(0,1,0) = 1 branch_3_A(0,1,1) = 1 branch_3_A(1,0,0) = 1 branch_3_A(1,0,1) = 1 branch_3_A(1,1,0) = 1 branch_3_A(1,1,1) = 1 leaf_1_A(0) = 1 leaf_1_A(1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 branch_0_A = 0 branch_1_A(0) = 1 branch_1_A(1) = 1 branch_2_A(0,0) = 0 branch_2_A(0,1) = 0 branch_2_A(1,0) = 0 branch_2_A(1,1) = 0 leaf_0_A = 0 mapbt_0_A = 1 mapbt_2_A(0,0) = 1 mapbt_2_A(0,1) = 1 mapbt_2_A(1,0) = 1 mapbt_2_A(1,1) = 1 app#!_A(0,0) = 1 app#!_A(0,1) = 1 app#!_A(1,0) = 1 app#!_A(1,1) = 1 mapbt_1!_A(0) = 1 mapbt_1!_A(1) = 1 mapbt_2#!_A(0,0) = 1 mapbt_2#!_A(0,1) = 1 mapbt_2#!_A(1,0) = 1 mapbt_2#!_A(1,1) = 1 branch_3!_A(0,0,0) = 0 branch_3!_A(0,0,1) = 0 branch_3!_A(0,1,0) = 0 branch_3!_A(0,1,1) = 0 branch_3!_A(1,0,0) = 0 branch_3!_A(1,0,1) = 0 branch_3!_A(1,1,0) = 0 branch_3!_A(1,1,1) = 0 leaf_1!_A(0) = 0 leaf_1!_A(1) = 0 app!_A(0,0) = 0 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 branch_0!_A = 0 branch_1!_A(0) = 0 branch_1!_A(1) = 0 branch_2!_A(0,0) = 1 branch_2!_A(0,1) = 0 branch_2!_A(1,0) = 0 branch_2!_A(1,1) = 1 leaf_0!_A = 1 mapbt_0!_A = 1 mapbt_2!_A(0,0) = 0 mapbt_2!_A(0,1) = 0 mapbt_2!_A(1,0) = 0 mapbt_2!_A(1,1) = 0 partial status: pi(app#) = [1] pi(mapbt_1) = [1] pi(mapbt_2#) = [1] pi(branch_3) = [3] pi(leaf_1) = [] pi(app) = [2] pi(branch_0) = [] pi(branch_1) = [1] pi(branch_2) = [] pi(leaf_0) = [] pi(mapbt_0) = [] pi(mapbt_2) = [2] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r) p2: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l) p3: mapbt_2#(f,branch_3(x,l,r)) -> app#(f,x) p4: mapbt_2#(f,leaf_1(x)) -> app#(f,x) and R consists of: r1: app(branch_0(),y) -> branch_1(y) r2: app(branch_1(x1),y) -> branch_2(x1,y) r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) r4: app(leaf_0(),y) -> leaf_1(y) r5: app(mapbt_0(),y) -> mapbt_1(y) r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y) r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,r) p2: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l) and R consists of: r1: app(branch_0(),y) -> branch_1(y) r2: app(branch_1(x1),y) -> branch_2(x1,y) r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) r4: app(leaf_0(),y) -> leaf_1(y) r5: app(mapbt_0(),y) -> mapbt_1(y) r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y) r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) 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: mapbt_2#_A(0,0) = 0 mapbt_2#_A(0,1) = 0 mapbt_2#_A(1,0) = 0 mapbt_2#_A(1,1) = 0 branch_3_A(0,0,0) = 1 branch_3_A(0,0,1) = 1 branch_3_A(0,1,0) = 1 branch_3_A(0,1,1) = 1 branch_3_A(1,0,0) = 1 branch_3_A(1,0,1) = 1 branch_3_A(1,1,0) = 1 branch_3_A(1,1,1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 branch_0_A = 0 branch_1_A(0) = 1 branch_1_A(1) = 1 branch_2_A(0,0) = 0 branch_2_A(0,1) = 0 branch_2_A(1,0) = 0 branch_2_A(1,1) = 0 leaf_0_A = 0 leaf_1_A(0) = 1 leaf_1_A(1) = 1 mapbt_0_A = 1 mapbt_1_A(0) = 0 mapbt_1_A(1) = 0 mapbt_2_A(0,0) = 1 mapbt_2_A(0,1) = 1 mapbt_2_A(1,0) = 1 mapbt_2_A(1,1) = 1 mapbt_2#!_A(0,0) = 1 mapbt_2#!_A(0,1) = 1 mapbt_2#!_A(1,0) = 0 mapbt_2#!_A(1,1) = 0 branch_3!_A(0,0,0) = 1 branch_3!_A(0,0,1) = 0 branch_3!_A(0,1,0) = 0 branch_3!_A(0,1,1) = 0 branch_3!_A(1,0,0) = 0 branch_3!_A(1,0,1) = 0 branch_3!_A(1,1,0) = 0 branch_3!_A(1,1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 branch_0!_A = 1 branch_1!_A(0) = 0 branch_1!_A(1) = 0 branch_2!_A(0,0) = 0 branch_2!_A(0,1) = 0 branch_2!_A(1,0) = 0 branch_2!_A(1,1) = 0 leaf_0!_A = 1 leaf_1!_A(0) = 0 leaf_1!_A(1) = 0 mapbt_0!_A = 0 mapbt_1!_A(0) = 0 mapbt_1!_A(1) = 0 mapbt_2!_A(0,0) = 0 mapbt_2!_A(0,1) = 0 mapbt_2!_A(1,0) = 0 mapbt_2!_A(1,1) = 0 partial status: pi(mapbt_2#) = [2] pi(branch_3) = [2, 3] pi(app) = [1, 2] pi(branch_0) = [] pi(branch_1) = [] pi(branch_2) = [2] pi(leaf_0) = [] pi(leaf_1) = [] pi(mapbt_0) = [] pi(mapbt_1) = [1] pi(mapbt_2) = [2] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l) and R consists of: r1: app(branch_0(),y) -> branch_1(y) r2: app(branch_1(x1),y) -> branch_2(x1,y) r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) r4: app(leaf_0(),y) -> leaf_1(y) r5: app(mapbt_0(),y) -> mapbt_1(y) r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y) r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: mapbt_2#(f,branch_3(x,l,r)) -> mapbt_2#(f,l) and R consists of: r1: app(branch_0(),y) -> branch_1(y) r2: app(branch_1(x1),y) -> branch_2(x1,y) r3: app(branch_2(x1,x2),y) -> branch_3(x1,x2,y) r4: app(leaf_0(),y) -> leaf_1(y) r5: app(mapbt_0(),y) -> mapbt_1(y) r6: app(mapbt_1(x1),y) -> mapbt_2(x1,y) r7: mapbt_2(f,leaf_1(x)) -> leaf_1(app(f,x)) r8: mapbt_2(f,branch_3(x,l,r)) -> branch_3(app(f,x),mapbt_2(f,l),mapbt_2(f,r)) 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: mapbt_2#_A(0,0) = 0 mapbt_2#_A(0,1) = 0 mapbt_2#_A(1,0) = 0 mapbt_2#_A(1,1) = 0 branch_3_A(0,0,0) = 1 branch_3_A(0,0,1) = 1 branch_3_A(0,1,0) = 1 branch_3_A(0,1,1) = 1 branch_3_A(1,0,0) = 1 branch_3_A(1,0,1) = 1 branch_3_A(1,1,0) = 1 branch_3_A(1,1,1) = 1 app_A(0,0) = 1 app_A(0,1) = 1 app_A(1,0) = 0 app_A(1,1) = 0 branch_0_A = 0 branch_1_A(0) = 1 branch_1_A(1) = 1 branch_2_A(0,0) = 0 branch_2_A(0,1) = 0 branch_2_A(1,0) = 0 branch_2_A(1,1) = 0 leaf_0_A = 0 leaf_1_A(0) = 1 leaf_1_A(1) = 1 mapbt_0_A = 1 mapbt_1_A(0) = 0 mapbt_1_A(1) = 0 mapbt_2_A(0,0) = 1 mapbt_2_A(0,1) = 1 mapbt_2_A(1,0) = 1 mapbt_2_A(1,1) = 1 mapbt_2#!_A(0,0) = 0 mapbt_2#!_A(0,1) = 0 mapbt_2#!_A(1,0) = 0 mapbt_2#!_A(1,1) = 0 branch_3!_A(0,0,0) = 0 branch_3!_A(0,0,1) = 0 branch_3!_A(0,1,0) = 0 branch_3!_A(0,1,1) = 0 branch_3!_A(1,0,0) = 0 branch_3!_A(1,0,1) = 0 branch_3!_A(1,1,0) = 0 branch_3!_A(1,1,1) = 0 app!_A(0,0) = 1 app!_A(0,1) = 1 app!_A(1,0) = 1 app!_A(1,1) = 1 branch_0!_A = 0 branch_1!_A(0) = 0 branch_1!_A(1) = 0 branch_2!_A(0,0) = 1 branch_2!_A(0,1) = 1 branch_2!_A(1,0) = 1 branch_2!_A(1,1) = 1 leaf_0!_A = 1 leaf_1!_A(0) = 0 leaf_1!_A(1) = 0 mapbt_0!_A = 0 mapbt_1!_A(0) = 0 mapbt_1!_A(1) = 0 mapbt_2!_A(0,0) = 1 mapbt_2!_A(0,1) = 1 mapbt_2!_A(1,0) = 0 mapbt_2!_A(1,1) = 1 partial status: pi(mapbt_2#) = [1, 2] pi(branch_3) = [2] pi(app) = [1, 2] pi(branch_0) = [] pi(branch_1) = [1] pi(branch_2) = [1, 2] pi(leaf_0) = [] pi(leaf_1) = [] pi(mapbt_0) = [] pi(mapbt_1) = [1] pi(mapbt_2) = [1, 2] The following rules are removed: p1 Then no dependency pair remains.