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synthesis:ltlformat [2011/04/19 17:30]
jobstman created
synthesis:ltlformat [2011/04/19 17:32]
jobstman
Line 18: Line 18:
 ===== Wring LTL-Syntax (used by Lily) ===== ===== Wring LTL-Syntax (used by Lily) =====
  
-FORMULA ​   ::= TERM {BINARYOP TERM} +FORMULA ​   ::= TERM {BINARYOP TERM}\\ 
-TERM       ::= {ATOM | (FORMULA) | UNARYOP FORMULA | TEMPORALOP (FORMULA)} +TERM       ::= {ATOM | (FORMULA) | UNARYOP FORMULA | TEMPORALOP (FORMULA)}\\ 
-BINARYOP ​  ::= * | + | ^ | -> | <-> | U | R | V +BINARYOP ​  ::= * | + | ^ | -> | <-> | U | R | V\\ 
-UNARYOP ​   ::= ! +UNARYOP ​   ::= !\\ 
-TEMPORALOP ::= G | F | X +TEMPORALOP ::= G | F | X\\ 
-ATOM       ::= VAR=VALUE +ATOM       ::= VAR=VALUE\\ 
-VAR        ::= \w+ +VAR        ::= \w+\\ 
-VALUE      ::= 0 | 1+VALUE      ::= 0 | 1\\
  
-Furthermore,​ Wring allows you to use the keyword "​define"​ to avoid writing the same formula multiple times+Furthermore,​ Wring allows you to use the keyword "​define"​ to avoid writing the same formula multiple times.\\
  
 Extensions in Lily: Extensions in Lily:
   * A formula can range over several lines. A semi-colon indicates the end of the formula.   * A formula can range over several lines. A semi-colon indicates the end of the formula.
   * Each formula can be prefixed with the keyword "​assume"​ or "​assert"​   * Each formula can be prefixed with the keyword "​assume"​ or "​assert"​
-  * In general, a semi-colon acts as conjunction between formulas unless the assume/​assert keywords are used. Then, the list of formulas corresponds to an implication between a conjunction of all formulas assumed and a conjunction of all formulas asserted. E.g., "assume G(a); assume G(b); assert G(c);" ​corresponds to "(G(a) * G(b)) -> G(b)".+  * In general, a semi-colon acts as conjunction between formulas unless the assume/​assert keywords are used. Then, the list of formulas corresponds to an implication between a conjunction of all formulas assumed and a conjunction of all formulas asserted. E.g.,\\ assume G(a=1);\\ assume G(b=1);\\ assert G(c=1);\\ corresponds to\\ (G(a=1) * G(b=1)) -> G(c=1)