LTS Termination Proof

by AProVE

Input

• Initial Location: l1, l3, l0, l2
• Transitions: (pre-variables and post-variables)

  l0	1	l1:		x1 = _xHAT0 ∧ x2 = _yHAT0 ∧ x1 = _xHATpost ∧ x2 = _yHATpost ∧ 1 ≤ _xHAT0 ∧ _yHAT1 = _xHAT0 ∧ _yHAT1 ≤ 0 ∧ _yHATpost = 1 + _yHAT1 ∧ _xHAT0 = _xHATpost
  l1	2	l0:		x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 = _x3 ∧ _x = _x2
  l2	3	l0:		x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 = _x7 ∧ _x4 = _x6
  l3	4	l2:		x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ _x9 = _x11 ∧ _x8 = _x10

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 }.

2.1.1 Transition Removal

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

2.1.2 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)