by T2Cert
0 | 0 | 1: | 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 0 | |
1 | 1 | 2: | 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ −1 + arg1 − 2⋅x6 ≤ 0 ∧ 1 − arg1 + 2⋅x6 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 | |
2 | 2 | 1: | 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ −1 + arg1 − 2⋅x8 ≤ 0 ∧ 1 − arg1 + 2⋅x8 ≤ 0 ∧ −1 + arg1 − 2⋅x8 ≤ 0 ∧ − arg1 + 2⋅x8 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 0 | |
1 | 3 | 2: | 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ arg1 − 2⋅x10 ≤ 0 ∧ − arg1 + 2⋅x10 ≤ 0 ∧ 1 − arg1 + x11 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0 | |
2 | 4 | 1: | 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ arg1 − 2⋅x14 ≤ 0 ∧ − arg1 + 2⋅x14 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ − arg1 + 2⋅x14 ≤ 0 ∧ −1 + arg1 − 2⋅x14 ≤ 0 ∧ −1 − 2⋅arg1P + arg1 ≤ 0 ∧ 2⋅arg1P − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 0 | |
3 | 5 | 0: | 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 0 |
The following invariants are asserted.
0: | TRUE |
1: | TRUE |
2: | 1 − arg1 ≤ 0 |
3: | TRUE |
The invariants are proved as follows.
0 | (0) | TRUE | ||
1 | (1) | TRUE | ||
2 | (2) | 1 − arg1 ≤ 0 | ||
3 | (3) | TRUE |
0 | 0 1 | |
1 | 1 2 | |
1 | 3 2 | |
2 | 2 1 | |
2 | 4 1 | |
3 | 5 0 |
1 | 6 | : | − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 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 |
We remove transitions
, using the following ranking functions, which are bounded by −11.3: | 0 |
0: | 0 |
1: | 0 |
2: | 0 |
: | −4 |
: | −5 |
: | −6 |
: | −6 |
: | −6 |
: | −6 |
7 | lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] |
lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] | |
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] ] | |
lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] | |
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] ] | |
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] ] |
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
9 : − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 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
The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.
7 : − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − x11 + x11 ≤ 0 ∧ x11 − x11 ≤ 0 ∧ − x10 + x10 ≤ 0 ∧ x10 − x10 ≤ 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
We consider subproblems for each of the 1 SCC(s) of the program graph.
Here we consider the SCC {
, , , }.We remove transitions
, , , using the following ranking functions, which are bounded by 3.: | 2 + 4⋅arg1 |
: | 4⋅arg1 |
: | 1 + 4⋅arg1 |
: | 3 + 4⋅arg1 |
7 | lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] |
9 | lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] ] |
lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0] , [0, 0, 0, 0, 0, 0, 4, 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, 4, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 4, 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, 4, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] | |
lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [4, 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] ] |
We remove transition 9 using the following ranking functions, which are bounded by −1.
: | 0 |
: | 0 |
: | −1 |
: | 1 |
7 | lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ] |
9 | 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] ] |
We remove transition 7 using the following ranking functions, which are bounded by −1.
: | 0 |
: | 0 |
: | −1 |
: | 0 |
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, 0, 0, 0, 0] ] |
We consider 1 subproblems corresponding to sets of cut-point transitions as follows.
There remain no cut-point transition to consider. Hence the cooperation termination is trivial.
T2Cert