# LTS Termination Proof

by T2Cert

## Input

Integer Transition System
• Initial Location: 2
• Transitions: (pre-variables and post-variables)  0 0 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg2 ≤ 0 ∧ − arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 2 − arg3P ≤ 0 ∧ −2 + arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 1 1 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg3 ≤ 0 ∧ 42 − x7 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ 1 − arg3P + arg3 ≤ 0 ∧ −1 + arg3P − arg3 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 1 2 1: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg3 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − x12 ≤ 0 ∧ − arg1P ≤ 0 ∧ −41 + x12 ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 0 ∧ 2 − arg3P + arg3 ≤ 0 ∧ −2 + arg3P − arg3 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 2 3 0: 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 1 4 1: − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0
and for every transition t, a duplicate t is considered.

### 2 Transition Removal

We remove transitions 0, 3 using the following ranking functions, which are bounded by −11.

 2: 0 0: 0 1: 0 2: −4 0: −5 1: −6 1_var_snapshot: −6 1*: −6
Hints:
 5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 1 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 2 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 0 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 3 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

1* 7 1: x7 + x7 ≤ 0x7x7 ≤ 0x12 + x12 ≤ 0x12x12 ≤ 0arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

1 5 1_var_snapshot: x7 + x7 ≤ 0x7x7 ≤ 0x12 + x12 ≤ 0x12x12 ≤ 0arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

### 5 SCC Decomposition

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

### 5.1 SCC Subproblem 1/1

Here we consider the SCC { 1, 1_var_snapshot, 1* }.

### 5.1.1 Transition Removal

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

 1: arg2 1_var_snapshot: arg2 1*: arg2
Hints:
 5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] ] 7 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] ] 1 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0] ] 2 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 5.1.2 Transition Removal

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

 1: 1 + 3⋅arg1 1_var_snapshot: 3⋅arg1 1*: 2 + 3⋅arg1
Hints:
 5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ] 7 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ] 1 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 5.1.3 Transition Removal

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

 1: 0 1_var_snapshot: −1 1*: 1
Hints:
 5 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] 7 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 5.1.4 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by −1.

 1: −1 1_var_snapshot: 0 1*: 0
Hints:
 7 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

### 5.1.5 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

### 5.1.5.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 4.

### 5.1.5.1.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

T2Cert

• version: 1.0