LTS Termination Proof

by AProVE

Input

• Initial Location: l5, l4, l1, l0, l2
• Transitions: (pre-variables and post-variables)

  l0	1	l1:		x1 = _xHAT0 ∧ x1 = _xHATpost ∧ _xHATpost = 1 ∧ _xHAT0 ≤ 0
  l0	2	l1:		x1 = _x ∧ x1 = _x1 ∧ _x1 = 1 + _x ∧ 1 ≤ _x
  l2	3	l3:		x1 = _x2 ∧ x1 = _x3 ∧ _x2 = _x3 ∧ 4 ≤ _x2
  l2	4	l0:		x1 = _x4 ∧ x1 = _x5 ∧ _x4 = _x5 ∧ 1 + _x4 ≤ 4
  l1	5	l2:		x1 = _x6 ∧ x1 = _x7 ∧ _x6 = _x7
  l4	6	l1:		x1 = _x8 ∧ x1 = _x9 ∧ _x10 = 5 ∧ _x9 = _x9
  l5	7	l4:		x1 = _x11 ∧ x1 = _x12 ∧ _x11 = _x12

Proof

1 Switch to Cooperation Termination Proof

2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { l1, l0, l2 }.

2.1.1 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

2.1.2 Transition Removal

We remove transitions 1, 5, 2 using the following ranking functions, which are bounded by −1.

2.1.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

Tool configuration

AProVE

• version: AProVE Commit ID: unknown
• strategy: Statistics for single proof: 100.00 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)