YES We show the termination of the TRS R: a__zeros() -> cons(|0|(),zeros()) a__U11(tt(),L) -> s(a__length(mark(L))) a__and(tt(),X) -> mark(X) a__isNat(|0|()) -> tt() a__isNat(length(V1)) -> a__isNatList(V1) a__isNat(s(V1)) -> a__isNat(V1) a__isNatIList(V) -> a__isNatList(V) a__isNatIList(zeros()) -> tt() a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2)) a__isNatList(nil()) -> tt() a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2)) a__length(nil()) -> |0|() a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L) mark(zeros()) -> a__zeros() mark(U11(X1,X2)) -> a__U11(mark(X1),X2) mark(length(X)) -> a__length(mark(X)) mark(and(X1,X2)) -> a__and(mark(X1),X2) mark(isNat(X)) -> a__isNat(X) mark(isNatList(X)) -> a__isNatList(X) mark(isNatIList(X)) -> a__isNatIList(X) mark(cons(X1,X2)) -> cons(mark(X1),X2) mark(|0|()) -> |0|() mark(tt()) -> tt() mark(s(X)) -> s(mark(X)) mark(nil()) -> nil() a__zeros() -> zeros() a__U11(X1,X2) -> U11(X1,X2) a__length(X) -> length(X) a__and(X1,X2) -> and(X1,X2) a__isNat(X) -> isNat(X) a__isNatList(X) -> isNatList(X) a__isNatIList(X) -> isNatIList(X) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__U11#(tt(),L) -> a__length#(mark(L)) p2: a__U11#(tt(),L) -> mark#(L) p3: a__and#(tt(),X) -> mark#(X) p4: a__isNat#(length(V1)) -> a__isNatList#(V1) p5: a__isNat#(s(V1)) -> a__isNat#(V1) p6: a__isNatIList#(V) -> a__isNatList#(V) p7: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2)) p8: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1) p9: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2)) p10: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1) p11: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L) p12: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N)) p13: a__length#(cons(N,L)) -> a__isNatList#(L) p14: mark#(zeros()) -> a__zeros#() p15: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2) p16: mark#(U11(X1,X2)) -> mark#(X1) p17: mark#(length(X)) -> a__length#(mark(X)) p18: mark#(length(X)) -> mark#(X) p19: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p20: mark#(and(X1,X2)) -> mark#(X1) p21: mark#(isNat(X)) -> a__isNat#(X) p22: mark#(isNatList(X)) -> a__isNatList#(X) p23: mark#(isNatIList(X)) -> a__isNatIList#(X) p24: mark#(cons(X1,X2)) -> mark#(X1) p25: mark#(s(X)) -> mark#(X) and R consists of: r1: a__zeros() -> cons(|0|(),zeros()) r2: a__U11(tt(),L) -> s(a__length(mark(L))) r3: a__and(tt(),X) -> mark(X) r4: a__isNat(|0|()) -> tt() r5: a__isNat(length(V1)) -> a__isNatList(V1) r6: a__isNat(s(V1)) -> a__isNat(V1) r7: a__isNatIList(V) -> a__isNatList(V) r8: a__isNatIList(zeros()) -> tt() r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2)) r10: a__isNatList(nil()) -> tt() r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2)) r12: a__length(nil()) -> |0|() r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L) r14: mark(zeros()) -> a__zeros() r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2) r16: mark(length(X)) -> a__length(mark(X)) r17: mark(and(X1,X2)) -> a__and(mark(X1),X2) r18: mark(isNat(X)) -> a__isNat(X) r19: mark(isNatList(X)) -> a__isNatList(X) r20: mark(isNatIList(X)) -> a__isNatIList(X) r21: mark(cons(X1,X2)) -> cons(mark(X1),X2) r22: mark(|0|()) -> |0|() r23: mark(tt()) -> tt() r24: mark(s(X)) -> s(mark(X)) r25: mark(nil()) -> nil() r26: a__zeros() -> zeros() r27: a__U11(X1,X2) -> U11(X1,X2) r28: a__length(X) -> length(X) r29: a__and(X1,X2) -> and(X1,X2) r30: a__isNat(X) -> isNat(X) r31: a__isNatList(X) -> isNatList(X) r32: a__isNatIList(X) -> isNatIList(X) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p15, p16, p17, p18, p19, p20, p21, p22, p23, p24, p25} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__U11#(tt(),L) -> a__length#(mark(L)) p2: a__length#(cons(N,L)) -> a__isNatList#(L) p3: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1) p4: a__isNat#(s(V1)) -> a__isNat#(V1) p5: a__isNat#(length(V1)) -> a__isNatList#(V1) p6: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2)) p7: a__and#(tt(),X) -> mark#(X) p8: mark#(s(X)) -> mark#(X) p9: mark#(cons(X1,X2)) -> mark#(X1) p10: mark#(isNatIList(X)) -> a__isNatIList#(X) p11: a__isNatIList#(cons(V1,V2)) -> a__isNat#(V1) p12: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2)) p13: a__isNatIList#(V) -> a__isNatList#(V) p14: mark#(isNatList(X)) -> a__isNatList#(X) p15: mark#(isNat(X)) -> a__isNat#(X) p16: mark#(and(X1,X2)) -> mark#(X1) p17: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p18: mark#(length(X)) -> mark#(X) p19: mark#(length(X)) -> a__length#(mark(X)) p20: a__length#(cons(N,L)) -> a__and#(a__isNatList(L),isNat(N)) p21: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L) p22: a__U11#(tt(),L) -> mark#(L) p23: mark#(U11(X1,X2)) -> mark#(X1) p24: mark#(U11(X1,X2)) -> a__U11#(mark(X1),X2) and R consists of: r1: a__zeros() -> cons(|0|(),zeros()) r2: a__U11(tt(),L) -> s(a__length(mark(L))) r3: a__and(tt(),X) -> mark(X) r4: a__isNat(|0|()) -> tt() r5: a__isNat(length(V1)) -> a__isNatList(V1) r6: a__isNat(s(V1)) -> a__isNat(V1) r7: a__isNatIList(V) -> a__isNatList(V) r8: a__isNatIList(zeros()) -> tt() r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2)) r10: a__isNatList(nil()) -> tt() r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2)) r12: a__length(nil()) -> |0|() r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L) r14: mark(zeros()) -> a__zeros() r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2) r16: mark(length(X)) -> a__length(mark(X)) r17: mark(and(X1,X2)) -> a__and(mark(X1),X2) r18: mark(isNat(X)) -> a__isNat(X) r19: mark(isNatList(X)) -> a__isNatList(X) r20: mark(isNatIList(X)) -> a__isNatIList(X) r21: mark(cons(X1,X2)) -> cons(mark(X1),X2) r22: mark(|0|()) -> |0|() r23: mark(tt()) -> tt() r24: mark(s(X)) -> s(mark(X)) r25: mark(nil()) -> nil() r26: a__zeros() -> zeros() r27: a__U11(X1,X2) -> U11(X1,X2) r28: a__length(X) -> length(X) r29: a__and(X1,X2) -> and(X1,X2) r30: a__isNat(X) -> isNat(X) r31: a__isNatList(X) -> isNatList(X) r32: a__isNatIList(X) -> isNatIList(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32 Take the reduction pair: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^2 order: standard order interpretations: a__U11#_A(x1,x2) = ((0,1),(0,1)) x2 + (2,0) tt_A() = (0,0) a__length#_A(x1) = ((1,0),(1,0)) x1 + (2,0) mark_A(x1) = ((0,1),(1,1)) x1 cons_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2 a__isNatList#_A(x1) = (0,0) a__isNat#_A(x1) = (0,0) s_A(x1) = x1 length_A(x1) = x1 + (1,3) a__and#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,1),(0,1)) x2 a__isNat_A(x1) = (0,0) isNatList_A(x1) = (0,0) mark#_A(x1) = ((0,1),(0,1)) x1 isNatIList_A(x1) = (1,2) a__isNatIList#_A(x1) = (2,2) isNat_A(x1) = (0,0) and_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2 a__isNatList_A(x1) = (0,0) a__and_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2 U11_A(x1,x2) = x1 + ((0,0),(1,1)) x2 + (1,3) a__zeros_A() = (1,1) |0|_A() = (0,0) zeros_A() = (0,1) a__U11_A(x1,x2) = x1 + ((0,1),(1,1)) x2 + (2,3) a__length_A(x1) = x1 + (2,3) a__isNatIList_A(x1) = (2,3) nil_A() = (1,2) 2. matrix interpretations: carrier: N^2 order: standard order interpretations: a__U11#_A(x1,x2) = (1,1) tt_A() = (0,0) a__length#_A(x1) = ((0,1),(0,0)) x1 + (1,1) mark_A(x1) = (0,0) cons_A(x1,x2) = (0,0) a__isNatList#_A(x1) = (0,0) a__isNat#_A(x1) = (0,0) s_A(x1) = (0,0) length_A(x1) = (1,2) a__and#_A(x1,x2) = (0,0) a__isNat_A(x1) = (0,0) isNatList_A(x1) = (0,0) mark#_A(x1) = (0,0) isNatIList_A(x1) = (0,0) a__isNatIList#_A(x1) = (0,0) isNat_A(x1) = (0,0) and_A(x1,x2) = (0,0) a__isNatList_A(x1) = (0,0) a__and_A(x1,x2) = (0,0) U11_A(x1,x2) = (2,0) a__zeros_A() = (0,0) |0|_A() = (0,0) zeros_A() = (1,1) a__U11_A(x1,x2) = (1,1) a__length_A(x1) = (2,1) a__isNatIList_A(x1) = (0,0) nil_A() = (0,1) The next rules are strictly ordered: p2, p11, p13, p18, p19, p20, p22, p23, p24 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: a__U11#(tt(),L) -> a__length#(mark(L)) p2: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1) p3: a__isNat#(s(V1)) -> a__isNat#(V1) p4: a__isNat#(length(V1)) -> a__isNatList#(V1) p5: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2)) p6: a__and#(tt(),X) -> mark#(X) p7: mark#(s(X)) -> mark#(X) p8: mark#(cons(X1,X2)) -> mark#(X1) p9: mark#(isNatIList(X)) -> a__isNatIList#(X) p10: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2)) p11: mark#(isNatList(X)) -> a__isNatList#(X) p12: mark#(isNat(X)) -> a__isNat#(X) p13: mark#(and(X1,X2)) -> mark#(X1) p14: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p15: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L) and R consists of: r1: a__zeros() -> cons(|0|(),zeros()) r2: a__U11(tt(),L) -> s(a__length(mark(L))) r3: a__and(tt(),X) -> mark(X) r4: a__isNat(|0|()) -> tt() r5: a__isNat(length(V1)) -> a__isNatList(V1) r6: a__isNat(s(V1)) -> a__isNat(V1) r7: a__isNatIList(V) -> a__isNatList(V) r8: a__isNatIList(zeros()) -> tt() r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2)) r10: a__isNatList(nil()) -> tt() r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2)) r12: a__length(nil()) -> |0|() r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L) r14: mark(zeros()) -> a__zeros() r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2) r16: mark(length(X)) -> a__length(mark(X)) r17: mark(and(X1,X2)) -> a__and(mark(X1),X2) r18: mark(isNat(X)) -> a__isNat(X) r19: mark(isNatList(X)) -> a__isNatList(X) r20: mark(isNatIList(X)) -> a__isNatIList(X) r21: mark(cons(X1,X2)) -> cons(mark(X1),X2) r22: mark(|0|()) -> |0|() r23: mark(tt()) -> tt() r24: mark(s(X)) -> s(mark(X)) r25: mark(nil()) -> nil() r26: a__zeros() -> zeros() r27: a__U11(X1,X2) -> U11(X1,X2) r28: a__length(X) -> length(X) r29: a__and(X1,X2) -> and(X1,X2) r30: a__isNat(X) -> isNat(X) r31: a__isNatList(X) -> isNatList(X) r32: a__isNatIList(X) -> isNatIList(X) The estimated dependency graph contains the following SCCs: {p1, p15} {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__U11#(tt(),L) -> a__length#(mark(L)) p2: a__length#(cons(N,L)) -> a__U11#(a__and(a__isNatList(L),isNat(N)),L) and R consists of: r1: a__zeros() -> cons(|0|(),zeros()) r2: a__U11(tt(),L) -> s(a__length(mark(L))) r3: a__and(tt(),X) -> mark(X) r4: a__isNat(|0|()) -> tt() r5: a__isNat(length(V1)) -> a__isNatList(V1) r6: a__isNat(s(V1)) -> a__isNat(V1) r7: a__isNatIList(V) -> a__isNatList(V) r8: a__isNatIList(zeros()) -> tt() r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2)) r10: a__isNatList(nil()) -> tt() r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2)) r12: a__length(nil()) -> |0|() r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L) r14: mark(zeros()) -> a__zeros() r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2) r16: mark(length(X)) -> a__length(mark(X)) r17: mark(and(X1,X2)) -> a__and(mark(X1),X2) r18: mark(isNat(X)) -> a__isNat(X) r19: mark(isNatList(X)) -> a__isNatList(X) r20: mark(isNatIList(X)) -> a__isNatIList(X) r21: mark(cons(X1,X2)) -> cons(mark(X1),X2) r22: mark(|0|()) -> |0|() r23: mark(tt()) -> tt() r24: mark(s(X)) -> s(mark(X)) r25: mark(nil()) -> nil() r26: a__zeros() -> zeros() r27: a__U11(X1,X2) -> U11(X1,X2) r28: a__length(X) -> length(X) r29: a__and(X1,X2) -> and(X1,X2) r30: a__isNat(X) -> isNat(X) r31: a__isNatList(X) -> isNatList(X) r32: a__isNatIList(X) -> isNatIList(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32 Take the reduction pair: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^2 order: standard order interpretations: a__U11#_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (2,0) tt_A() = (16,1) a__length#_A(x1) = ((1,1),(0,0)) x1 mark_A(x1) = x1 + (17,0) cons_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (15,0) a__and_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (1,0) a__isNatList_A(x1) = x1 + (7,1) isNat_A(x1) = x1 + (1,1) a__zeros_A() = (17,1) |0|_A() = (0,1) zeros_A() = (0,1) a__U11_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (4,1) s_A(x1) = x1 + (1,1) a__length_A(x1) = ((1,1),(0,1)) x1 + (1,1) a__isNat_A(x1) = x1 + (17,1) length_A(x1) = ((1,1),(0,1)) x1 + (1,1) a__isNatIList_A(x1) = x1 + (17,1) isNatIList_A(x1) = x1 + (1,1) nil_A() = (10,1) U11_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (4,1) isNatList_A(x1) = x1 + (1,1) and_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (1,0) 2. matrix interpretations: carrier: N^2 order: standard order interpretations: a__U11#_A(x1,x2) = ((1,1),(0,0)) x2 tt_A() = (5,2) a__length#_A(x1) = (1,0) mark_A(x1) = ((0,1),(0,0)) x1 + (5,4) cons_A(x1,x2) = (0,1) a__and_A(x1,x2) = (4,3) a__isNatList_A(x1) = (3,1) isNat_A(x1) = ((0,0),(1,1)) x1 + (5,2) a__zeros_A() = (1,0) |0|_A() = (1,1) zeros_A() = (2,1) a__U11_A(x1,x2) = ((0,1),(0,0)) x1 + (0,3) s_A(x1) = (3,4) a__length_A(x1) = (2,0) a__isNat_A(x1) = (4,1) length_A(x1) = (1,0) a__isNatIList_A(x1) = (2,0) isNatIList_A(x1) = (3,1) nil_A() = (1,1) U11_A(x1,x2) = ((0,1),(0,0)) x1 isNatList_A(x1) = ((1,1),(0,1)) x1 + (4,2) and_A(x1,x2) = (1,0) The next rules are strictly ordered: p1, p2 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: a__isNatList#(cons(V1,V2)) -> a__isNat#(V1) p2: a__isNat#(length(V1)) -> a__isNatList#(V1) p3: a__isNatList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatList(V2)) p4: a__and#(tt(),X) -> mark#(X) p5: mark#(and(X1,X2)) -> a__and#(mark(X1),X2) p6: mark#(and(X1,X2)) -> mark#(X1) p7: mark#(isNat(X)) -> a__isNat#(X) p8: a__isNat#(s(V1)) -> a__isNat#(V1) p9: mark#(isNatList(X)) -> a__isNatList#(X) p10: mark#(isNatIList(X)) -> a__isNatIList#(X) p11: a__isNatIList#(cons(V1,V2)) -> a__and#(a__isNat(V1),isNatIList(V2)) p12: mark#(cons(X1,X2)) -> mark#(X1) p13: mark#(s(X)) -> mark#(X) and R consists of: r1: a__zeros() -> cons(|0|(),zeros()) r2: a__U11(tt(),L) -> s(a__length(mark(L))) r3: a__and(tt(),X) -> mark(X) r4: a__isNat(|0|()) -> tt() r5: a__isNat(length(V1)) -> a__isNatList(V1) r6: a__isNat(s(V1)) -> a__isNat(V1) r7: a__isNatIList(V) -> a__isNatList(V) r8: a__isNatIList(zeros()) -> tt() r9: a__isNatIList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatIList(V2)) r10: a__isNatList(nil()) -> tt() r11: a__isNatList(cons(V1,V2)) -> a__and(a__isNat(V1),isNatList(V2)) r12: a__length(nil()) -> |0|() r13: a__length(cons(N,L)) -> a__U11(a__and(a__isNatList(L),isNat(N)),L) r14: mark(zeros()) -> a__zeros() r15: mark(U11(X1,X2)) -> a__U11(mark(X1),X2) r16: mark(length(X)) -> a__length(mark(X)) r17: mark(and(X1,X2)) -> a__and(mark(X1),X2) r18: mark(isNat(X)) -> a__isNat(X) r19: mark(isNatList(X)) -> a__isNatList(X) r20: mark(isNatIList(X)) -> a__isNatIList(X) r21: mark(cons(X1,X2)) -> cons(mark(X1),X2) r22: mark(|0|()) -> |0|() r23: mark(tt()) -> tt() r24: mark(s(X)) -> s(mark(X)) r25: mark(nil()) -> nil() r26: a__zeros() -> zeros() r27: a__U11(X1,X2) -> U11(X1,X2) r28: a__length(X) -> length(X) r29: a__and(X1,X2) -> and(X1,X2) r30: a__isNat(X) -> isNat(X) r31: a__isNatList(X) -> isNatList(X) r32: a__isNatIList(X) -> isNatIList(X) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32 Take the reduction pair: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^2 order: standard order interpretations: a__isNatList#_A(x1) = x1 + (26,0) cons_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (19,0) a__isNat#_A(x1) = x1 + (26,0) length_A(x1) = ((1,1),(0,1)) x1 + (1,1) a__and#_A(x1,x2) = x1 + x2 a__isNat_A(x1) = x1 + (24,1) isNatList_A(x1) = x1 + (5,0) tt_A() = (23,0) mark#_A(x1) = x1 + (22,0) and_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,0) mark_A(x1) = x1 + (21,0) isNat_A(x1) = x1 + (4,1) s_A(x1) = x1 + (1,1) isNatIList_A(x1) = x1 + (1,0) a__isNatIList#_A(x1) = x1 + (22,0) a__zeros_A() = (21,1) |0|_A() = (0,1) zeros_A() = (1,1) a__U11_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (1,1) a__length_A(x1) = ((1,1),(0,1)) x1 + (1,1) a__and_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,0) a__isNatIList_A(x1) = x1 + (22,0) a__isNatList_A(x1) = x1 + (12,0) nil_A() = (11,0) U11_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (1,1) 2. matrix interpretations: carrier: N^2 order: standard order interpretations: a__isNatList#_A(x1) = ((1,1),(0,0)) x1 + (2,1) cons_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,18) a__isNat#_A(x1) = x1 length_A(x1) = x1 + (1,1) a__and#_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (0,1) a__isNat_A(x1) = (1,1) isNatList_A(x1) = ((0,1),(1,1)) x1 tt_A() = (4,0) mark#_A(x1) = ((1,1),(0,0)) x1 + (3,1) and_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (1,4) mark_A(x1) = ((0,1),(0,1)) x1 + (6,6) isNat_A(x1) = ((0,0),(1,1)) x1 + (2,1) s_A(x1) = x1 + (2,3) isNatIList_A(x1) = ((1,1),(1,1)) x1 + (1,15) a__isNatIList#_A(x1) = (18,2) a__zeros_A() = (10,24) |0|_A() = (5,2) zeros_A() = (1,3) a__U11_A(x1,x2) = x1 + (3,10) a__length_A(x1) = x1 + (2,1) a__and_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (3,4) a__isNatIList_A(x1) = ((1,1),(0,0)) x1 + (0,21) a__isNatList_A(x1) = ((0,1),(0,1)) x1 + (5,2) nil_A() = (1,0) U11_A(x1,x2) = x1 + (3,10) The next rules are strictly ordered: p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13 We remove them from the problem. Then no dependency pair remains.