LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l6, l1, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = ___const_100HAT0 ∧ x2 = ___const_200HAT0 ∧ x3 = _iHAT0 ∧ x4 = _jHAT0 ∧ x5 = _y4HAT0 ∧ x6 = _y6HAT0 ∧ x7 = _y8HAT0 ∧ x1 = ___const_100HATpost ∧ x2 = ___const_200HATpost ∧ x3 = _iHATpost ∧ x4 = _jHATpost ∧ x5 = _y4HATpost ∧ x6 = _y6HATpost ∧ x7 = _y8HATpost ∧ _y8HAT0 = _y8HATpost ∧ _y6HAT0 = _y6HATpost ∧ _y4HAT0 = _y4HATpost ∧ _iHAT0 = _iHATpost ∧ ___const_200HAT0 = ___const_200HATpost ∧ ___const_100HAT0 = ___const_100HATpost ∧ _jHATpost = ___const_100HAT0 ∧ ___const_100HAT0 ≤ _iHAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x7 = _x6 ∧ x1 = _x7 ∧ x2 = _x8 ∧ x3 = _x9 ∧ x4 = _x10 ∧ x5 = _x11 ∧ x6 = _x12 ∧ x7 = _x13 ∧ _x6 = _x13 ∧ _x3 = _x10 ∧ _x1 = _x8 ∧ _x = _x7 ∧ _x9 = 1 + _x2 ∧ _x12 = _x2 ∧ _x11 = _x2 ∧ 1 + _x2 ≤ _x
l2	3	l0:	x1 = _x14 ∧ x2 = _x15 ∧ x3 = _x16 ∧ x4 = _x17 ∧ x5 = _x18 ∧ x6 = _x19 ∧ x7 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ x4 = _x24 ∧ x5 = _x25 ∧ x6 = _x26 ∧ x7 = _x27 ∧ _x20 = _x27 ∧ _x19 = _x26 ∧ _x18 = _x25 ∧ _x17 = _x24 ∧ _x16 = _x23 ∧ _x15 = _x22 ∧ _x14 = _x21
l3	4	l4:	x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ x5 = _x32 ∧ x6 = _x33 ∧ x7 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ x6 = _x40 ∧ x7 = _x41 ∧ _x34 = _x41 ∧ _x33 = _x40 ∧ _x32 = _x39 ∧ _x31 = _x38 ∧ _x30 = _x37 ∧ _x29 = _x36 ∧ _x28 = _x35 ∧ _x29 ≤ _x31
l3	5	l1:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x4 = _x45 ∧ x5 = _x46 ∧ x6 = _x47 ∧ x7 = _x48 ∧ x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x5 = _x53 ∧ x6 = _x54 ∧ x7 = _x55 ∧ _x47 = _x54 ∧ _x46 = _x53 ∧ _x44 = _x51 ∧ _x43 = _x50 ∧ _x42 = _x49 ∧ _x52 = 1 + _x45 ∧ _x55 = _x45 ∧ 1 + _x45 ≤ _x43
l1	6	l3:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x5 = _x60 ∧ x6 = _x61 ∧ x7 = _x62 ∧ x1 = _x63 ∧ x2 = _x64 ∧ x3 = _x65 ∧ x4 = _x66 ∧ x5 = _x67 ∧ x6 = _x68 ∧ x7 = _x69 ∧ _x62 = _x69 ∧ _x61 = _x68 ∧ _x60 = _x67 ∧ _x59 = _x66 ∧ _x58 = _x65 ∧ _x57 = _x64 ∧ _x56 = _x63
l5	7	l2:	x1 = _x70 ∧ x2 = _x71 ∧ x3 = _x72 ∧ x4 = _x73 ∧ x5 = _x74 ∧ x6 = _x75 ∧ x7 = _x76 ∧ x1 = _x77 ∧ x2 = _x78 ∧ x3 = _x79 ∧ x4 = _x80 ∧ x5 = _x81 ∧ x6 = _x82 ∧ x7 = _x83 ∧ _x76 = _x83 ∧ _x75 = _x82 ∧ _x74 = _x81 ∧ _x73 = _x80 ∧ _x71 = _x78 ∧ _x70 = _x77 ∧ _x79 = 0
l6	8	l5:	x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x4 = _x87 ∧ x5 = _x88 ∧ x6 = _x89 ∧ x7 = _x90 ∧ x1 = _x91 ∧ x2 = _x92 ∧ x3 = _x93 ∧ x4 = _x94 ∧ x5 = _x95 ∧ x6 = _x96 ∧ x7 = _x97 ∧ _x90 = _x97 ∧ _x89 = _x96 ∧ _x88 = _x95 ∧ _x87 = _x94 ∧ _x86 = _x93 ∧ _x85 = _x92 ∧ _x84 = _x91

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7

and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

Here we consider the SCC { l0, l2 }.

2.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

l0:	2⋅x1 − 2⋅x3
l2:	2⋅x1 − 2⋅x3 + 1

2.1.2 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

l2:	0
l0:	−1

2.1.3 Trivial Cooperation Program

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

2.2 SCC Subproblem 2/2

Here we consider the SCC { l1, l3 }.

2.2.1 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

l1:	−1 + x2 − x4
l3:	−1 + x2 − x4

2.2.2 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

l1:	0
l3:	−1

2.2.3 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 % (10 real / 0 unknown / 0 assumptions / 10 total proof steps)