Certification Problem

Input (TPDB TRS_Standard/CiME_04/log2)

The rewrite relation of the following TRS is considered.

0(#) # (1)
+(#,x) x (2)
+(x,#) x (3)
+(0(x),0(y)) 0(+(x,y)) (4)
+(0(x),1(y)) 1(+(x,y)) (5)
+(1(x),0(y)) 1(+(x,y)) (6)
+(1(x),1(y)) 0(+(+(x,y),1(#))) (7)
+(+(x,y),z) +(x,+(y,z)) (8)
-(#,x) # (9)
-(x,#) x (10)
-(0(x),0(y)) 0(-(x,y)) (11)
-(0(x),1(y)) 1(-(-(x,y),1(#))) (12)
-(1(x),0(y)) 1(-(x,y)) (13)
-(1(x),1(y)) 0(-(x,y)) (14)
not(true) false (15)
not(false) true (16)
if(true,x,y) x (17)
if(false,x,y) y (18)
ge(0(x),0(y)) ge(x,y) (19)
ge(0(x),1(y)) not(ge(y,x)) (20)
ge(1(x),0(y)) ge(x,y) (21)
ge(1(x),1(y)) ge(x,y) (22)
ge(x,#) true (23)
ge(#,0(x)) ge(#,x) (24)
ge(#,1(x)) false (25)
log(x) -(log'(x),1(#)) (26)
log'(#) # (27)
log'(1(x)) +(log'(x),1(#)) (28)
log'(0(x)) if(ge(x,1(#)),+(log'(x),1(#)),#) (29)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
log'#(0(x)) ge#(x,1(#)) (30)
+#(0(x),0(y)) 0#(+(x,y)) (31)
-#(1(x),0(y)) -#(x,y) (32)
-#(1(x),1(y)) -#(x,y) (33)
ge#(#,0(x)) ge#(#,x) (34)
-#(0(x),1(y)) -#(x,y) (35)
log#(x) -#(log'(x),1(#)) (36)
ge#(0(x),0(y)) ge#(x,y) (37)
-#(1(x),1(y)) 0#(-(x,y)) (38)
ge#(0(x),1(y)) not#(ge(y,x)) (39)
+#(0(x),0(y)) +#(x,y) (40)
log#(x) log'#(x) (41)
log'#(1(x)) +#(log'(x),1(#)) (42)
-#(0(x),0(y)) 0#(-(x,y)) (43)
-#(0(x),1(y)) -#(-(x,y),1(#)) (44)
+#(1(x),0(y)) +#(x,y) (45)
+#(+(x,y),z) +#(y,z) (46)
ge#(1(x),1(y)) ge#(x,y) (47)
+#(+(x,y),z) +#(x,+(y,z)) (48)
+#(1(x),1(y)) +#(+(x,y),1(#)) (49)
+#(0(x),1(y)) +#(x,y) (50)
ge#(1(x),0(y)) ge#(x,y) (51)
log'#(1(x)) log'#(x) (52)
+#(1(x),1(y)) 0#(+(+(x,y),1(#))) (53)
log'#(0(x)) +#(log'(x),1(#)) (54)
log'#(0(x)) if#(ge(x,1(#)),+(log'(x),1(#)),#) (55)
-#(0(x),0(y)) -#(x,y) (56)
ge#(0(x),1(y)) ge#(y,x) (57)
log'#(0(x)) log'#(x) (58)
+#(1(x),1(y)) +#(x,y) (59)

1.1 Dependency Graph Processor

The dependency pairs are split into 5 components.