YES We show the termination of the TRS R: U11(tt(),N) -> activate(N) U21(tt(),M,N) -> s(plus(activate(N),activate(M))) and(tt(),X) -> activate(X) isNat(n__0()) -> tt() isNat(n__plus(V1,V2)) -> and(isNat(activate(V1)),n__isNat(activate(V2))) isNat(n__s(V1)) -> isNat(activate(V1)) plus(N,|0|()) -> U11(isNat(N),N) plus(N,s(M)) -> U21(and(isNat(M),n__isNat(N)),M,N) |0|() -> n__0() plus(X1,X2) -> n__plus(X1,X2) isNat(X) -> n__isNat(X) s(X) -> n__s(X) activate(n__0()) -> |0|() activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2)) activate(n__isNat(X)) -> isNat(X) activate(n__s(X)) -> s(activate(X)) activate(X) -> X -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: U11#(tt(),N) -> activate#(N) p2: U21#(tt(),M,N) -> s#(plus(activate(N),activate(M))) p3: U21#(tt(),M,N) -> plus#(activate(N),activate(M)) p4: U21#(tt(),M,N) -> activate#(N) p5: U21#(tt(),M,N) -> activate#(M) p6: and#(tt(),X) -> activate#(X) p7: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2))) p8: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1)) p9: isNat#(n__plus(V1,V2)) -> activate#(V1) p10: isNat#(n__plus(V1,V2)) -> activate#(V2) p11: isNat#(n__s(V1)) -> isNat#(activate(V1)) p12: isNat#(n__s(V1)) -> activate#(V1) p13: plus#(N,|0|()) -> U11#(isNat(N),N) p14: plus#(N,|0|()) -> isNat#(N) p15: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N) p16: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N)) p17: plus#(N,s(M)) -> isNat#(M) p18: activate#(n__0()) -> |0|#() p19: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2)) p20: activate#(n__plus(X1,X2)) -> activate#(X1) p21: activate#(n__plus(X1,X2)) -> activate#(X2) p22: activate#(n__isNat(X)) -> isNat#(X) p23: activate#(n__s(X)) -> s#(activate(X)) p24: activate#(n__s(X)) -> activate#(X) and R consists of: r1: U11(tt(),N) -> activate(N) r2: U21(tt(),M,N) -> s(plus(activate(N),activate(M))) r3: and(tt(),X) -> activate(X) r4: isNat(n__0()) -> tt() r5: isNat(n__plus(V1,V2)) -> and(isNat(activate(V1)),n__isNat(activate(V2))) r6: isNat(n__s(V1)) -> isNat(activate(V1)) r7: plus(N,|0|()) -> U11(isNat(N),N) r8: plus(N,s(M)) -> U21(and(isNat(M),n__isNat(N)),M,N) r9: |0|() -> n__0() r10: plus(X1,X2) -> n__plus(X1,X2) r11: isNat(X) -> n__isNat(X) r12: s(X) -> n__s(X) r13: activate(n__0()) -> |0|() r14: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2)) r15: activate(n__isNat(X)) -> isNat(X) r16: activate(n__s(X)) -> s(activate(X)) r17: activate(X) -> X The estimated dependency graph contains the following SCCs: {p1, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p19, p20, p21, p22, p24} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: U11#(tt(),N) -> activate#(N) p2: activate#(n__s(X)) -> activate#(X) p3: activate#(n__isNat(X)) -> isNat#(X) p4: isNat#(n__s(V1)) -> activate#(V1) p5: activate#(n__plus(X1,X2)) -> activate#(X2) p6: activate#(n__plus(X1,X2)) -> activate#(X1) p7: activate#(n__plus(X1,X2)) -> plus#(activate(X1),activate(X2)) p8: plus#(N,s(M)) -> isNat#(M) p9: isNat#(n__s(V1)) -> isNat#(activate(V1)) p10: isNat#(n__plus(V1,V2)) -> activate#(V2) p11: isNat#(n__plus(V1,V2)) -> activate#(V1) p12: isNat#(n__plus(V1,V2)) -> isNat#(activate(V1)) p13: isNat#(n__plus(V1,V2)) -> and#(isNat(activate(V1)),n__isNat(activate(V2))) p14: and#(tt(),X) -> activate#(X) p15: plus#(N,s(M)) -> and#(isNat(M),n__isNat(N)) p16: plus#(N,s(M)) -> U21#(and(isNat(M),n__isNat(N)),M,N) p17: U21#(tt(),M,N) -> activate#(M) p18: U21#(tt(),M,N) -> activate#(N) p19: U21#(tt(),M,N) -> plus#(activate(N),activate(M)) p20: plus#(N,|0|()) -> isNat#(N) p21: plus#(N,|0|()) -> U11#(isNat(N),N) and R consists of: r1: U11(tt(),N) -> activate(N) r2: U21(tt(),M,N) -> s(plus(activate(N),activate(M))) r3: and(tt(),X) -> activate(X) r4: isNat(n__0()) -> tt() r5: isNat(n__plus(V1,V2)) -> and(isNat(activate(V1)),n__isNat(activate(V2))) r6: isNat(n__s(V1)) -> isNat(activate(V1)) r7: plus(N,|0|()) -> U11(isNat(N),N) r8: plus(N,s(M)) -> U21(and(isNat(M),n__isNat(N)),M,N) r9: |0|() -> n__0() r10: plus(X1,X2) -> n__plus(X1,X2) r11: isNat(X) -> n__isNat(X) r12: s(X) -> n__s(X) r13: activate(n__0()) -> |0|() r14: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2)) r15: activate(n__isNat(X)) -> isNat(X) r16: activate(n__s(X)) -> s(activate(X)) r17: activate(X) -> 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 Take the reduction pair: matrix interpretations: carrier: N^2 order: lexicographic order interpretations: U11#_A(x1,x2) = ((1,0),(1,0)) x2 + (1,2) tt_A() = (0,1) activate#_A(x1) = ((1,0),(1,0)) x1 + (1,2) n__s_A(x1) = x1 + (9,0) n__isNat_A(x1) = x1 isNat#_A(x1) = ((1,0),(1,0)) x1 + (1,2) n__plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (7,0) plus#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 activate_A(x1) = ((1,0),(1,1)) x1 + (0,5) s_A(x1) = x1 + (9,1) and#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (8,3) isNat_A(x1) = ((1,0),(1,0)) x1 + (0,1) U21#_A(x1,x2,x3) = ((1,0),(1,0)) x2 + x3 + (2,6) and_A(x1,x2) = ((1,0),(1,0)) x2 + (1,9) |0|_A() = (1,1) U11_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,4) U21_A(x1,x2,x3) = ((1,0),(1,0)) x2 + ((1,0),(1,1)) x3 + (16,2) plus_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (7,1) n__0_A() = (1,0) The next rules are strictly ordered: p2, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17, p18, p19 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: U11#(tt(),N) -> activate#(N) p2: activate#(n__isNat(X)) -> isNat#(X) p3: plus#(N,|0|()) -> isNat#(N) p4: plus#(N,|0|()) -> U11#(isNat(N),N) and R consists of: r1: U11(tt(),N) -> activate(N) r2: U21(tt(),M,N) -> s(plus(activate(N),activate(M))) r3: and(tt(),X) -> activate(X) r4: isNat(n__0()) -> tt() r5: isNat(n__plus(V1,V2)) -> and(isNat(activate(V1)),n__isNat(activate(V2))) r6: isNat(n__s(V1)) -> isNat(activate(V1)) r7: plus(N,|0|()) -> U11(isNat(N),N) r8: plus(N,s(M)) -> U21(and(isNat(M),n__isNat(N)),M,N) r9: |0|() -> n__0() r10: plus(X1,X2) -> n__plus(X1,X2) r11: isNat(X) -> n__isNat(X) r12: s(X) -> n__s(X) r13: activate(n__0()) -> |0|() r14: activate(n__plus(X1,X2)) -> plus(activate(X1),activate(X2)) r15: activate(n__isNat(X)) -> isNat(X) r16: activate(n__s(X)) -> s(activate(X)) r17: activate(X) -> X The estimated dependency graph contains the following SCCs: (no SCCs)