NO Initial ITS Start location: [1] Program variables: x 0: [1] -> [2] : x'=nondet, T, cost: 1 1: [2] -> [3] : -10+x <= 0, cost: 1 7: [2] -> [7] : -10+x > 0, cost: 1 2: [3] -> [5] : -6+x > 0, cost: 1 5: [3] -> [4] : -6+x <= 0, cost: 1 6: [4] -> [2] : T, cost: 1 3: [5] -> [6] : x'=2+x, T, cost: 1 4: [6] -> [4] : T, cost: 1 Chained Linear Paths Start location: [1] Program variables: x 0: [1] -> [2] : x'=nondet, T, cost: 1 1: [2] -> [3] : -10+x <= 0, cost: 1 7: [2] -> [7] : -10+x > 0, cost: 1 5: [3] -> [4] : -6+x <= 0, cost: 1 9: [3] -> [4] : x'=2+x, -6+x > 0, cost: 1 6: [4] -> [2] : T, cost: 1 Eliminating location [5] by chaining: Applied chaining First rule: [3] -> [5] : -6+x > 0, cost: 1 Second rule: [5] -> [6] : x'=2+x, T, cost: 1 New rule: [3] -> [6] : x'=2+x, -6+x > 0, cost: 1 Applied deletion Removed the following rules: 2 3 Eliminating location [6] by chaining: Applied chaining First rule: [3] -> [6] : x'=2+x, -6+x > 0, cost: 1 Second rule: [6] -> [4] : T, cost: 1 New rule: [3] -> [4] : x'=2+x, -6+x > 0, cost: 1 Applied deletion Removed the following rules: 4 8 Step with 0 Trace 0[T] Blocked [{}, {}] Step with 7 Trace 0[T], 7[(-10+x > 0)] Blocked [{}, {}, {}] Backtrack Trace 0[T] Blocked [{}, {7[T]}] Step with 1 Trace 0[T], 1[(-10+x <= 0)] Blocked [{}, {7[T]}, {}] Step with 5 Trace 0[T], 1[(-10+x <= 0)], 5[(-6+x <= 0)] Blocked [{}, {7[T]}, {}, {}] Step with 6 Trace 0[T], 1[(-10+x <= 0)], 5[(-6+x <= 0)], 6[T] Blocked [{}, {7[T]}, {}, {}, {}] Nonterm Start location: [1] Program variables: x 0: [1] -> [2] : x'=nondet, T, cost: 1 1: [2] -> [3] : -10+x <= 0, cost: 1 7: [2] -> [7] : -10+x > 0, cost: 1 10: [2] -> LoAT_sink : (10-x >= 0 /\ 6-x >= 0 /\ -1+n >= 0), cost: NONTERM 5: [3] -> [4] : -6+x <= 0, cost: 1 9: [3] -> [4] : x'=2+x, -6+x > 0, cost: 1 6: [4] -> [2] : T, cost: 1 Certificate of Non-Termination Original rule: [2] -> [2] : (-6+x <= 0 /\ -10+x <= 0), cost: 1 New rule: [2] -> LoAT_sink : (10-x >= 0 /\ 6-x >= 0 /\ -1+n >= 0), cost: NONTERM 10-x >= 0 [0]: monotonic increase yields 10-x >= 0 6-x >= 0 [0]: monotonic increase yields 6-x >= 0 Replacement map: {10-x >= 0 -> 10-x >= 0, 6-x >= 0 -> 6-x >= 0} Step with 10 Trace 0[T], 10[(10-x >= 0 /\ 6-x >= 0 /\ -1+n >= 0)] Blocked [{}, {7[T]}, {10[T]}] Refute Counterexample [ x=6 ] 0 [ x=x ] 10 NO Build SHA: a05f16bf13df659c382799650051f91bf6828c7b