Coq: (a :: L1) = (b :: L2) ⇒ a = b ∧ L1 = L2?
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
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
add a comment |
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
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
add a comment |
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
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
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
coq
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
asked Nov 10 '18 at 19:20
Harry QuHarry Qu
206
206
add a comment |
add a comment |
1 Answer
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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).
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
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1 Answer
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1 Answer
1
active
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active
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active
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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).
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
add a comment |
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).
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
add a comment |
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).
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
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).
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
answered Nov 10 '18 at 19:36
ejgallegoejgallego
5,3541925
5,3541925
add a comment |
add a comment |
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