0.00/0.00	MAYBE
0.00/0.00	
0.00/0.00	
0.00/0.00	Succeeded in reading "/export/starexec/sandbox/benchmark/theBenchmark.trs".
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      or(0,x) -> x
0.00/0.00	      or(x,0) -> x
0.00/0.00	      or(1,x) -> 1
0.00/0.00	      or(x,1) -> 1
0.00/0.00	      or(x,not(x)) -> 1
0.00/0.00	      or(not(x),x) -> 1
0.00/0.00	      and(0,x) -> 0
0.00/0.00	      and(x,0) -> 0
0.00/0.00	      and(1,x) -> x
0.00/0.00	      and(x,1) -> x
0.00/0.00	      and(x,not(x)) -> 0
0.00/0.00	      and(not(x),x) -> 0
0.00/0.00	      not(1) -> 0
0.00/0.00	      not(0) -> 1
0.00/0.00	      implies(x,y) -> 1 | not(x) == 1
0.00/0.00	      implies(x,y) -> 1 | y == 1
0.00/0.00	      implies(x,y) -> 0 | x == 1, y == 0
0.00/0.00	      f(x) -> f(0) | implies(x,0) == y, implies(x,y) == z, implies(z,0) == 1
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.1007/978-3-319-13770-4_3 [151] Example 1 submitted by: Raul Gutierrez and Salvador Lucas)
0.00/0.00	
0.00/0.00	No "->="-rules.
0.00/0.00	
0.00/0.00	Decomposed conditions if possible.
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      or(0,x) -> x
0.00/0.00	      or(x,0) -> x
0.00/0.00	      or(1,x) -> 1
0.00/0.00	      or(x,1) -> 1
0.00/0.00	      or(x,not(x)) -> 1
0.00/0.00	      or(not(x),x) -> 1
0.00/0.00	      and(0,x) -> 0
0.00/0.00	      and(x,0) -> 0
0.00/0.00	      and(1,x) -> x
0.00/0.00	      and(x,1) -> x
0.00/0.00	      and(x,not(x)) -> 0
0.00/0.00	      and(not(x),x) -> 0
0.00/0.00	      not(1) -> 0
0.00/0.00	      not(0) -> 1
0.00/0.00	      implies(x,y) -> 1 | not(x) == 1
0.00/0.00	      implies(x,y) -> 1 | y == 1
0.00/0.00	      implies(x,y) -> 0 | x == 1, y == 0
0.00/0.00	      f(x) -> f(0) | implies(x,0) == y, implies(x,y) == z, implies(z,0) == 1
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.1007/978-3-319-13770-4_3 [151] Example 1 submitted by: Raul Gutierrez and Salvador Lucas)
0.00/0.00	
0.00/0.00	Removed infeasible rules as much as possible.
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      or(0,x) -> x
0.00/0.00	      or(x,0) -> x
0.00/0.00	      or(1,x) -> 1
0.00/0.00	      or(x,1) -> 1
0.00/0.00	      or(x,not(x)) -> 1
0.00/0.00	      or(not(x),x) -> 1
0.00/0.00	      and(0,x) -> 0
0.00/0.00	      and(x,0) -> 0
0.00/0.00	      and(1,x) -> x
0.00/0.00	      and(x,1) -> x
0.00/0.00	      and(x,not(x)) -> 0
0.00/0.00	      and(not(x),x) -> 0
0.00/0.00	      not(1) -> 0
0.00/0.00	      not(0) -> 1
0.00/0.00	      implies(x,y) -> 1 | not(x) == 1
0.00/0.00	      implies(x,y) -> 1 | y == 1
0.00/0.00	      implies(x,y) -> 0 | x == 1, y == 0
0.00/0.00	      f(x) -> f(0) | implies(x,0) == y, implies(x,y) == z, implies(z,0) == 1
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.1007/978-3-319-13770-4_3 [151] Example 1 submitted by: Raul Gutierrez and Salvador Lucas)
0.00/0.00	
0.00/0.00	Try to disprove confluence of the following (C)TRS:
0.00/0.00	    (CONDITIONTYPE ORIENTED)
0.00/0.00	    (VAR x y z)
0.00/0.00	    (RULES
0.00/0.00	      or(0,x) -> x
0.00/0.00	      or(x,0) -> x
0.00/0.00	      or(1,x) -> 1
0.00/0.00	      or(x,1) -> 1
0.00/0.00	      or(x,not(x)) -> 1
0.00/0.00	      or(not(x),x) -> 1
0.00/0.00	      and(0,x) -> 0
0.00/0.00	      and(x,0) -> 0
0.00/0.00	      and(1,x) -> x
0.00/0.00	      and(x,1) -> x
0.00/0.00	      and(x,not(x)) -> 0
0.00/0.00	      and(not(x),x) -> 0
0.00/0.00	      not(1) -> 0
0.00/0.00	      not(0) -> 1
0.00/0.00	      implies(x,y) -> 1 | not(x) == 1
0.00/0.00	      implies(x,y) -> 1 | y == 1
0.00/0.00	      implies(x,y) -> 0 | x == 1, y == 0
0.00/0.00	      f(x) -> f(0) | implies(x,0) == y, implies(x,y) == z, implies(z,0) == 1
0.00/0.00	    )
0.00/0.00	    (COMMENT doi:10.1007/978-3-319-13770-4_3 [151] Example 1 submitted by: Raul Gutierrez and Salvador Lucas)
0.00/0.00	
0.00/0.00	Failed either to apply SR and U for normal 1CTRSs to the above CTRS or to prove confluence of any converted TRSs.
0.00/0.00	
0.00/0.00	Try to apply SR and U for 3DCTRSs to the above CTRS.
0.00/0.00	
0.00/0.00	Succeeded in applying U for 3DCTRSs to the above CTRS.
0.00/0.00	U(R) =
0.00/0.00	    (VAR x1 x2 x3)
0.00/0.00	    (RULES
0.00/0.00	      or(0,x1) -> x1
0.00/0.00	      or(x1,0) -> x1
0.00/0.00	      or(1,x1) -> 1
0.00/0.00	      or(x1,1) -> 1
0.00/0.00	      or(x1,not(x1)) -> 1
0.00/0.00	      or(not(x1),x1) -> 1
0.00/0.00	      and(0,x1) -> 0
0.00/0.00	      and(x1,0) -> 0
0.00/0.00	      and(1,x1) -> x1
0.00/0.00	      and(x1,1) -> x1
0.00/0.00	      and(x1,not(x1)) -> 0
0.00/0.00	      and(not(x1),x1) -> 0
0.00/0.00	      not(1) -> 0
0.00/0.00	      not(0) -> 1
0.00/0.00	      implies(x1,x2) -> u1(not(x1),x1,x2)
0.00/0.00	      u1(1,x1,x2) -> 1
0.00/0.00	      implies(x1,x2) -> u2(x2,x1,x2)
0.00/0.00	      u2(1,x1,x2) -> 1
0.00/0.00	      implies(x1,x2) -> u3(x1,x1,x2)
0.00/0.00	      u3(1,x1,x2) -> u4(x2,x1,x2)
0.00/0.00	      u4(0,x1,x2) -> 0
0.00/0.00	      f(x1) -> u5(implies(x1,0),x1)
0.00/0.00	      u5(x2,x1) -> u6(implies(x1,x2),x2,x1)
0.00/0.00	      u6(x3,x2,x1) -> u7(implies(x3,0),x3,x2,x1)
0.00/0.00	      u7(1,x3,x2,x1) -> f(0)
0.00/0.00	    )
0.00/0.00	
0.00/0.00	U for 3DCTRSs is sound for the above CTRS.
0.00/0.00	
0.00/0.00	Failed to prove confluence of U(R).
0.00/0.00	
0.00/0.00	Try to prove operational termination of R, i.e., termination of U(R).
0.00/0.00	
0.00/0.00	Failed to prove operational termination of R.
0.00/0.00	
0.00/0.00	MAYBE
0.00/0.00	EOF