LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l7, l6, l1, l8, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _i2HAT0 ∧ x2 = _size1010HAT0 ∧ x3 = _size1HAT0 ∧ x4 = _size77HAT0 ∧ x5 = _tmp1111HAT0 ∧ x6 = _tmp88HAT0 ∧ x1 = _i2HATpost ∧ x2 = _size1010HATpost ∧ x3 = _size1HATpost ∧ x4 = _size77HATpost ∧ x5 = _tmp1111HATpost ∧ x6 = _tmp88HATpost ∧ _tmp88HAT0 = _tmp88HATpost ∧ _tmp1111HAT0 = _tmp1111HATpost ∧ _size77HAT0 = _size77HATpost ∧ _size1010HAT0 = _size1010HATpost ∧ _size1HAT0 = _size1HATpost ∧ _i2HAT0 = _i2HATpost
l2	2	l3:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x4 = _x9 ∧ x5 = _x10 ∧ x6 = _x11 ∧ _x5 = _x11 ∧ _x4 = _x10 ∧ _x3 = _x9 ∧ _x1 = _x7 ∧ _x2 = _x8 ∧ _x = _x6 ∧ _x2 ≤ _x
l2	3	l4:	x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ x5 = _x16 ∧ x6 = _x17 ∧ x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x4 = _x21 ∧ x5 = _x22 ∧ x6 = _x23 ∧ _x17 = _x23 ∧ _x16 = _x22 ∧ _x15 = _x21 ∧ _x13 = _x19 ∧ _x14 = _x20 ∧ _x18 = 1 + _x12 ∧ 1 + _x12 ≤ _x14
l5	4	l6:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x5 = _x28 ∧ x6 = _x29 ∧ x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x6 = _x35 ∧ _x29 = _x35 ∧ _x28 = _x34 ∧ _x27 = _x33 ∧ _x25 = _x31 ∧ _x26 = _x32 ∧ _x24 = _x30
l6	5	l4:	x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ x5 = _x40 ∧ x6 = _x41 ∧ x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x4 = _x45 ∧ x5 = _x46 ∧ x6 = _x47 ∧ _x41 = _x47 ∧ _x40 = _x46 ∧ _x39 = _x45 ∧ _x37 = _x43 ∧ _x38 = _x44 ∧ _x42 = 0 ∧ _x38 ≤ _x36
l6	6	l5:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x5 = _x52 ∧ x6 = _x53 ∧ x1 = _x54 ∧ x2 = _x55 ∧ x3 = _x56 ∧ x4 = _x57 ∧ x5 = _x58 ∧ x6 = _x59 ∧ _x53 = _x59 ∧ _x52 = _x58 ∧ _x51 = _x57 ∧ _x49 = _x55 ∧ _x50 = _x56 ∧ _x54 = 1 + _x48 ∧ 1 + _x48 ≤ _x50
l4	7	l2:	x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ x5 = _x64 ∧ x6 = _x65 ∧ x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x4 = _x69 ∧ x5 = _x70 ∧ x6 = _x71 ∧ _x65 = _x71 ∧ _x64 = _x70 ∧ _x63 = _x69 ∧ _x61 = _x67 ∧ _x62 = _x68 ∧ _x60 = _x66
l1	8	l5:	x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x4 = _x75 ∧ x5 = _x76 ∧ x6 = _x77 ∧ x1 = _x78 ∧ x2 = _x79 ∧ x3 = _x80 ∧ x4 = _x81 ∧ x5 = _x82 ∧ x6 = _x83 ∧ _x77 = _x83 ∧ _x76 = _x82 ∧ _x75 = _x81 ∧ _x73 = _x79 ∧ _x74 = _x80 ∧ _x78 = 0 ∧ _x74 ≤ _x72
l1	9	l0:	x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x4 = _x87 ∧ x5 = _x88 ∧ x6 = _x89 ∧ x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x4 = _x93 ∧ x5 = _x94 ∧ x6 = _x95 ∧ _x89 = _x95 ∧ _x88 = _x94 ∧ _x87 = _x93 ∧ _x85 = _x91 ∧ _x86 = _x92 ∧ _x90 = 1 + _x84 ∧ 1 + _x84 ≤ _x86
l7	10	l0:	x1 = _x96 ∧ x2 = _x97 ∧ x3 = _x98 ∧ x4 = _x99 ∧ x5 = _x100 ∧ x6 = _x101 ∧ x1 = _x102 ∧ x2 = _x103 ∧ x3 = _x104 ∧ x4 = _x105 ∧ x5 = _x106 ∧ x6 = _x107 ∧ _x102 = 0 ∧ _x106 = _x106 ∧ _x103 = _x104 ∧ _x107 = _x107 ∧ _x105 = _x104 ∧ _x104 = 10
l8	11	l7:	x1 = _x108 ∧ x2 = _x109 ∧ x3 = _x110 ∧ x4 = _x111 ∧ x5 = _x112 ∧ x6 = _x113 ∧ x1 = _x114 ∧ x2 = _x115 ∧ x3 = _x116 ∧ x4 = _x117 ∧ x5 = _x118 ∧ x6 = _x119 ∧ _x113 = _x119 ∧ _x112 = _x118 ∧ _x111 = _x117 ∧ _x109 = _x115 ∧ _x110 = _x116 ∧ _x108 = _x114

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
l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l7	l7	l7:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l8	l8	l8:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6

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

2 SCC Decomposition

We consider subproblems for each of the 3 SCC(s) of the program graph.

2.1 SCC Subproblem 1/3

Here we consider the SCC { l1, l0 }.

2.1.1 Transition Removal

We remove transition 9 using the following ranking functions, which are bounded by 0.

l0:	−1 − x1 + x3
l1:	−1 − x1 + x3

2.1.2 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by 0.

l0:	0
l1:	−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 2/3

Here we consider the SCC { l5, l6 }.

2.2.1 Transition Removal

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

l5:	−1 − x1 + x3
l6:	−1 − x1 + x3

2.2.2 Transition Removal

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

l5:	0
l6:	−1

2.2.3 Trivial Cooperation Program

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

2.3 SCC Subproblem 3/3

Here we consider the SCC { l4, l2 }.

2.3.1 Transition Removal

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

l4:	−1 − x1 + x3
l2:	−1 − x1 + x3

2.3.2 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by 0.

l4:	0
l2:	−1

2.3.3 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 % (11 real / 0 unknown / 0 assumptions / 11 total proof steps)