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?
coq
asked Nov 10 at 19:20
Harry Qu
205
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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?
coq
asked Nov 10 at 19:20
Harry Qu
205
add a comment |
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?
coq
asked Nov 10 at 19:20
Harry Qu
205
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 at 19:20
Harry Qu
205
asked Nov 10 at 19:20
Harry Qu
205
asked Nov 10 at 19:20
Harry Qu
205
asked Nov 10 at 19:20
Harry Qu
205
asked Nov 10 at 19:20
Harry Qu
205
205
add a comment |
add a comment |
1 Answer
1
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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).
answered Nov 10 at 19:36
ejgallego
5,2541825
add a comment |
1 Answer
1
active
oldest
votes
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).
answered Nov 10 at 19:36
ejgallego
5,2541825
add a comment |
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).
answered Nov 10 at 19:36
ejgallego
5,2541825
add a comment |
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).
answered Nov 10 at 19:36
ejgallego
5,2541825
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 at 19:36
ejgallego
5,2541825
answered Nov 10 at 19:36
ejgallego
5,2541825
answered Nov 10 at 19:36
ejgallego
5,2541825
answered Nov 10 at 19:36
ejgallego
5,2541825
5,2541825
add a comment |
add a comment |
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