Coq: (a :: L1) = (b :: L2) ⇒ a = b ∧ L1 = L2?











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This statement seems obvious to me, unless I'm overlooking some counterexample, but I couldn't find anything in the Coq lists library that does this. Is there a command that does something to this effect?










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    up vote
    1
    down vote

    favorite












    This statement seems obvious to me, unless I'm overlooking some counterexample, but I couldn't find anything in the Coq lists library that does this. Is there a command that does something to this effect?










    share|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      This statement seems obvious to me, unless I'm overlooking some counterexample, but I couldn't find anything in the Coq lists library that does this. Is there a command that does something to this effect?










      share|improve this question













      This statement seems obvious to me, unless I'm overlooking some counterexample, but I couldn't find anything in the Coq lists library that does this. Is there a command that does something to this effect?







      coq






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      asked Nov 10 at 19:20









      Harry Qu

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          This can be usually derived using the injection tactic. A version of the lemma for rewriting can be found in math-comp:



          eqseq_cons (T : eqType) (x1 x2 : T) (s1 s2 : seq T) :
          (x1 :: s1 == x2 :: s2) = (x1 == x2) && (s1 == s2).





          share|improve this answer





















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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            This can be usually derived using the injection tactic. A version of the lemma for rewriting can be found in math-comp:



            eqseq_cons (T : eqType) (x1 x2 : T) (s1 s2 : seq T) :
            (x1 :: s1 == x2 :: s2) = (x1 == x2) && (s1 == s2).





            share|improve this answer

























              up vote
              3
              down vote



              accepted










              This can be usually derived using the injection tactic. A version of the lemma for rewriting can be found in math-comp:



              eqseq_cons (T : eqType) (x1 x2 : T) (s1 s2 : seq T) :
              (x1 :: s1 == x2 :: s2) = (x1 == x2) && (s1 == s2).





              share|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                This can be usually derived using the injection tactic. A version of the lemma for rewriting can be found in math-comp:



                eqseq_cons (T : eqType) (x1 x2 : T) (s1 s2 : seq T) :
                (x1 :: s1 == x2 :: s2) = (x1 == x2) && (s1 == s2).





                share|improve this answer












                This can be usually derived using the injection tactic. A version of the lemma for rewriting can be found in math-comp:



                eqseq_cons (T : eqType) (x1 x2 : T) (s1 s2 : seq T) :
                (x1 :: s1 == x2 :: s2) = (x1 == x2) && (s1 == s2).






                share|improve this answer












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                answered Nov 10 at 19:36









                ejgallego

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