A non-inductive proof for Euler's Formula?












2












$begingroup$


Theorem: The number of vertices $|V|$, edges $|E|$, and faces $|F|$ in an arbitrary connected planar graph are related by the formula $$|V|+|F|-|E|=2$$



Proof Attempt:



(For acyclic planar graphs) Let $G(V,E)$ be an acyclic graph. For any line-subgraph $L=(V',E')$ of $G$, we can easily verify that $|V'|=2$, $|E'|=1$, and $|F'|=1$ where $F'$ is the face of a line graph. Thus, $|V'|+|F'|-|E'|=2$. As we construct $G$ from $L$ by adding edges, note that for each edge we add, we add a vertex so for any number of edges $ninBbb{N}$ we add, we also add $n$ vertices. Hence, $(|V'|+n)+|F'|-(|E'|+n)=|V|+|F|-|E|=2$.



(For planar graphs with cyclic subgraphs) The smallest planar graph with a cyclic subgraph is $C_3$(a triangle). Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$. We assert that to add an edge to $C_3$, we either




  1. Add a point $v^*$ on an existing edge $l$ which 'splits' $l$ to two different edges then connect $v^*$ to an existing vertex with a new edge; this adds 2 edges, 1 vertex, and 1 face or


  2. Connect two existing vertices, not connected by an edge, with a new edge; this adds a face or


  3. Add 2 new vertices with each of them on an existing edge, then connect the two new vertices with a new edge; this adds 2 vertices, 1 face, and 3 edges.



Nevertheless, observe that method (1) and (3) above is a sequential combination of two very primitive steps:




  • Add a vertex to an existing edge.


  • Connect two existing vertices, not connected by an edge, with a
    new edge. (This is method (2) above.)



Note:




  • Adding a point to an existing edge implies $|V|+1$, $|E|+1$, and $|F|+0$.

  • Method 2 implies $|V|+0$, $|E|+1$, and $|F|+1$.

  • Adding a point outside any edge forces us to connect this new vertex with an existing vertex by a new edge; this implies $|V|+1$, $|E|+1$, and $|F|+0$.


Hence, the form of numerical parameter variations occurring every time we construct a new connected planar graph is of the following:




  • $(|V|+n)+|F|-(|E|+n)$

  • $|V|+(|F|+m)-(|E|+m)$


Therefore, $|V|+|F|-|E|=2$.










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Theorem: The number of vertices $|V|$, edges $|E|$, and faces $|F|$ in an arbitrary connected planar graph are related by the formula $$|V|+|F|-|E|=2$$



    Proof Attempt:



    (For acyclic planar graphs) Let $G(V,E)$ be an acyclic graph. For any line-subgraph $L=(V',E')$ of $G$, we can easily verify that $|V'|=2$, $|E'|=1$, and $|F'|=1$ where $F'$ is the face of a line graph. Thus, $|V'|+|F'|-|E'|=2$. As we construct $G$ from $L$ by adding edges, note that for each edge we add, we add a vertex so for any number of edges $ninBbb{N}$ we add, we also add $n$ vertices. Hence, $(|V'|+n)+|F'|-(|E'|+n)=|V|+|F|-|E|=2$.



    (For planar graphs with cyclic subgraphs) The smallest planar graph with a cyclic subgraph is $C_3$(a triangle). Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$. We assert that to add an edge to $C_3$, we either




    1. Add a point $v^*$ on an existing edge $l$ which 'splits' $l$ to two different edges then connect $v^*$ to an existing vertex with a new edge; this adds 2 edges, 1 vertex, and 1 face or


    2. Connect two existing vertices, not connected by an edge, with a new edge; this adds a face or


    3. Add 2 new vertices with each of them on an existing edge, then connect the two new vertices with a new edge; this adds 2 vertices, 1 face, and 3 edges.



    Nevertheless, observe that method (1) and (3) above is a sequential combination of two very primitive steps:




    • Add a vertex to an existing edge.


    • Connect two existing vertices, not connected by an edge, with a
      new edge. (This is method (2) above.)



    Note:




    • Adding a point to an existing edge implies $|V|+1$, $|E|+1$, and $|F|+0$.

    • Method 2 implies $|V|+0$, $|E|+1$, and $|F|+1$.

    • Adding a point outside any edge forces us to connect this new vertex with an existing vertex by a new edge; this implies $|V|+1$, $|E|+1$, and $|F|+0$.


    Hence, the form of numerical parameter variations occurring every time we construct a new connected planar graph is of the following:




    • $(|V|+n)+|F|-(|E|+n)$

    • $|V|+(|F|+m)-(|E|+m)$


    Therefore, $|V|+|F|-|E|=2$.










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Theorem: The number of vertices $|V|$, edges $|E|$, and faces $|F|$ in an arbitrary connected planar graph are related by the formula $$|V|+|F|-|E|=2$$



      Proof Attempt:



      (For acyclic planar graphs) Let $G(V,E)$ be an acyclic graph. For any line-subgraph $L=(V',E')$ of $G$, we can easily verify that $|V'|=2$, $|E'|=1$, and $|F'|=1$ where $F'$ is the face of a line graph. Thus, $|V'|+|F'|-|E'|=2$. As we construct $G$ from $L$ by adding edges, note that for each edge we add, we add a vertex so for any number of edges $ninBbb{N}$ we add, we also add $n$ vertices. Hence, $(|V'|+n)+|F'|-(|E'|+n)=|V|+|F|-|E|=2$.



      (For planar graphs with cyclic subgraphs) The smallest planar graph with a cyclic subgraph is $C_3$(a triangle). Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$. We assert that to add an edge to $C_3$, we either




      1. Add a point $v^*$ on an existing edge $l$ which 'splits' $l$ to two different edges then connect $v^*$ to an existing vertex with a new edge; this adds 2 edges, 1 vertex, and 1 face or


      2. Connect two existing vertices, not connected by an edge, with a new edge; this adds a face or


      3. Add 2 new vertices with each of them on an existing edge, then connect the two new vertices with a new edge; this adds 2 vertices, 1 face, and 3 edges.



      Nevertheless, observe that method (1) and (3) above is a sequential combination of two very primitive steps:




      • Add a vertex to an existing edge.


      • Connect two existing vertices, not connected by an edge, with a
        new edge. (This is method (2) above.)



      Note:




      • Adding a point to an existing edge implies $|V|+1$, $|E|+1$, and $|F|+0$.

      • Method 2 implies $|V|+0$, $|E|+1$, and $|F|+1$.

      • Adding a point outside any edge forces us to connect this new vertex with an existing vertex by a new edge; this implies $|V|+1$, $|E|+1$, and $|F|+0$.


      Hence, the form of numerical parameter variations occurring every time we construct a new connected planar graph is of the following:




      • $(|V|+n)+|F|-(|E|+n)$

      • $|V|+(|F|+m)-(|E|+m)$


      Therefore, $|V|+|F|-|E|=2$.










      share|cite|improve this question











      $endgroup$




      Theorem: The number of vertices $|V|$, edges $|E|$, and faces $|F|$ in an arbitrary connected planar graph are related by the formula $$|V|+|F|-|E|=2$$



      Proof Attempt:



      (For acyclic planar graphs) Let $G(V,E)$ be an acyclic graph. For any line-subgraph $L=(V',E')$ of $G$, we can easily verify that $|V'|=2$, $|E'|=1$, and $|F'|=1$ where $F'$ is the face of a line graph. Thus, $|V'|+|F'|-|E'|=2$. As we construct $G$ from $L$ by adding edges, note that for each edge we add, we add a vertex so for any number of edges $ninBbb{N}$ we add, we also add $n$ vertices. Hence, $(|V'|+n)+|F'|-(|E'|+n)=|V|+|F|-|E|=2$.



      (For planar graphs with cyclic subgraphs) The smallest planar graph with a cyclic subgraph is $C_3$(a triangle). Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$. We assert that to add an edge to $C_3$, we either




      1. Add a point $v^*$ on an existing edge $l$ which 'splits' $l$ to two different edges then connect $v^*$ to an existing vertex with a new edge; this adds 2 edges, 1 vertex, and 1 face or


      2. Connect two existing vertices, not connected by an edge, with a new edge; this adds a face or


      3. Add 2 new vertices with each of them on an existing edge, then connect the two new vertices with a new edge; this adds 2 vertices, 1 face, and 3 edges.



      Nevertheless, observe that method (1) and (3) above is a sequential combination of two very primitive steps:




      • Add a vertex to an existing edge.


      • Connect two existing vertices, not connected by an edge, with a
        new edge. (This is method (2) above.)



      Note:




      • Adding a point to an existing edge implies $|V|+1$, $|E|+1$, and $|F|+0$.

      • Method 2 implies $|V|+0$, $|E|+1$, and $|F|+1$.

      • Adding a point outside any edge forces us to connect this new vertex with an existing vertex by a new edge; this implies $|V|+1$, $|E|+1$, and $|F|+0$.


      Hence, the form of numerical parameter variations occurring every time we construct a new connected planar graph is of the following:




      • $(|V|+n)+|F|-(|E|+n)$

      • $|V|+(|F|+m)-(|E|+m)$


      Therefore, $|V|+|F|-|E|=2$.







      proof-verification graph-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 1 hour ago







      TheLast Cipher

















      asked 1 hour ago









      TheLast CipherTheLast Cipher

      742715




      742715






















          1 Answer
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          $begingroup$

          Your proof is not a non-inductive proof. The steps where you construct $G$ from a smaller graph by adding edges, and check that $|V|-|E|+|F|=2$ still holds because the left-hand side does not change, are the inductive steps of your proof.



          This is fine! Induction is great. But you are falling into the classic "induction trap" that everyone falls into when writing induction proofs about graphs.



          Whenever you write an inductive step that "goes up", starting with a graph for which your result holds and making it bigger, you set yourself the goal of showing that every graph can be obtained by growing your base case in this way. You are not careful about this:




          • In the cyclic case, you write "Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$". This should set off alarm bells. Whenever you are justifying an intuitively clear statement with "clearly", you haven't written a proof.

          • In the acyclic case, you haven't even stated the claim that your proof needs to work: that every tree can be built by repeatedly adding leaves to a line-graph. (Do you mean a path graph? A line graph is something else.)


          We can avoid the problem of proving these hard-to-prove statements by writing an inductive step that "goes down" instead. Here, we assume that the result holds for all small graphs, and take a slightly larger graph; then, delete a vertex or edge from it, reducing it to a graph that your inductive hypothesis applies to.



          For example, when proving Euler's formula for acyclic graphs, you could first argue (in your favorite way) that any such graph contains a vertex of degree $1$, then delete that vertex together with the edge out of it. This reduces the graph to a smaller one for which you have already verified Euler's formula.



          Here, there is no worry that we've gotten all of the graphs we're considering in this case, because we started with an arbitrary such graph.



          Aside from this, your strategy looks good. You should just rethink your arguments in terms of removing an edge or vertex, rather than adding them.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! This is very informative.
            $endgroup$
            – TheLast Cipher
            37 mins ago












          Your Answer








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          $begingroup$

          Your proof is not a non-inductive proof. The steps where you construct $G$ from a smaller graph by adding edges, and check that $|V|-|E|+|F|=2$ still holds because the left-hand side does not change, are the inductive steps of your proof.



          This is fine! Induction is great. But you are falling into the classic "induction trap" that everyone falls into when writing induction proofs about graphs.



          Whenever you write an inductive step that "goes up", starting with a graph for which your result holds and making it bigger, you set yourself the goal of showing that every graph can be obtained by growing your base case in this way. You are not careful about this:




          • In the cyclic case, you write "Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$". This should set off alarm bells. Whenever you are justifying an intuitively clear statement with "clearly", you haven't written a proof.

          • In the acyclic case, you haven't even stated the claim that your proof needs to work: that every tree can be built by repeatedly adding leaves to a line-graph. (Do you mean a path graph? A line graph is something else.)


          We can avoid the problem of proving these hard-to-prove statements by writing an inductive step that "goes down" instead. Here, we assume that the result holds for all small graphs, and take a slightly larger graph; then, delete a vertex or edge from it, reducing it to a graph that your inductive hypothesis applies to.



          For example, when proving Euler's formula for acyclic graphs, you could first argue (in your favorite way) that any such graph contains a vertex of degree $1$, then delete that vertex together with the edge out of it. This reduces the graph to a smaller one for which you have already verified Euler's formula.



          Here, there is no worry that we've gotten all of the graphs we're considering in this case, because we started with an arbitrary such graph.



          Aside from this, your strategy looks good. You should just rethink your arguments in terms of removing an edge or vertex, rather than adding them.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! This is very informative.
            $endgroup$
            – TheLast Cipher
            37 mins ago
















          4












          $begingroup$

          Your proof is not a non-inductive proof. The steps where you construct $G$ from a smaller graph by adding edges, and check that $|V|-|E|+|F|=2$ still holds because the left-hand side does not change, are the inductive steps of your proof.



          This is fine! Induction is great. But you are falling into the classic "induction trap" that everyone falls into when writing induction proofs about graphs.



          Whenever you write an inductive step that "goes up", starting with a graph for which your result holds and making it bigger, you set yourself the goal of showing that every graph can be obtained by growing your base case in this way. You are not careful about this:




          • In the cyclic case, you write "Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$". This should set off alarm bells. Whenever you are justifying an intuitively clear statement with "clearly", you haven't written a proof.

          • In the acyclic case, you haven't even stated the claim that your proof needs to work: that every tree can be built by repeatedly adding leaves to a line-graph. (Do you mean a path graph? A line graph is something else.)


          We can avoid the problem of proving these hard-to-prove statements by writing an inductive step that "goes down" instead. Here, we assume that the result holds for all small graphs, and take a slightly larger graph; then, delete a vertex or edge from it, reducing it to a graph that your inductive hypothesis applies to.



          For example, when proving Euler's formula for acyclic graphs, you could first argue (in your favorite way) that any such graph contains a vertex of degree $1$, then delete that vertex together with the edge out of it. This reduces the graph to a smaller one for which you have already verified Euler's formula.



          Here, there is no worry that we've gotten all of the graphs we're considering in this case, because we started with an arbitrary such graph.



          Aside from this, your strategy looks good. You should just rethink your arguments in terms of removing an edge or vertex, rather than adding them.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! This is very informative.
            $endgroup$
            – TheLast Cipher
            37 mins ago














          4












          4








          4





          $begingroup$

          Your proof is not a non-inductive proof. The steps where you construct $G$ from a smaller graph by adding edges, and check that $|V|-|E|+|F|=2$ still holds because the left-hand side does not change, are the inductive steps of your proof.



          This is fine! Induction is great. But you are falling into the classic "induction trap" that everyone falls into when writing induction proofs about graphs.



          Whenever you write an inductive step that "goes up", starting with a graph for which your result holds and making it bigger, you set yourself the goal of showing that every graph can be obtained by growing your base case in this way. You are not careful about this:




          • In the cyclic case, you write "Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$". This should set off alarm bells. Whenever you are justifying an intuitively clear statement with "clearly", you haven't written a proof.

          • In the acyclic case, you haven't even stated the claim that your proof needs to work: that every tree can be built by repeatedly adding leaves to a line-graph. (Do you mean a path graph? A line graph is something else.)


          We can avoid the problem of proving these hard-to-prove statements by writing an inductive step that "goes down" instead. Here, we assume that the result holds for all small graphs, and take a slightly larger graph; then, delete a vertex or edge from it, reducing it to a graph that your inductive hypothesis applies to.



          For example, when proving Euler's formula for acyclic graphs, you could first argue (in your favorite way) that any such graph contains a vertex of degree $1$, then delete that vertex together with the edge out of it. This reduces the graph to a smaller one for which you have already verified Euler's formula.



          Here, there is no worry that we've gotten all of the graphs we're considering in this case, because we started with an arbitrary such graph.



          Aside from this, your strategy looks good. You should just rethink your arguments in terms of removing an edge or vertex, rather than adding them.






          share|cite|improve this answer









          $endgroup$



          Your proof is not a non-inductive proof. The steps where you construct $G$ from a smaller graph by adding edges, and check that $|V|-|E|+|F|=2$ still holds because the left-hand side does not change, are the inductive steps of your proof.



          This is fine! Induction is great. But you are falling into the classic "induction trap" that everyone falls into when writing induction proofs about graphs.



          Whenever you write an inductive step that "goes up", starting with a graph for which your result holds and making it bigger, you set yourself the goal of showing that every graph can be obtained by growing your base case in this way. You are not careful about this:




          • In the cyclic case, you write "Clearly, any planar graph with a cyclic subgraph can be constructed by adding more points and edges to $C_3$". This should set off alarm bells. Whenever you are justifying an intuitively clear statement with "clearly", you haven't written a proof.

          • In the acyclic case, you haven't even stated the claim that your proof needs to work: that every tree can be built by repeatedly adding leaves to a line-graph. (Do you mean a path graph? A line graph is something else.)


          We can avoid the problem of proving these hard-to-prove statements by writing an inductive step that "goes down" instead. Here, we assume that the result holds for all small graphs, and take a slightly larger graph; then, delete a vertex or edge from it, reducing it to a graph that your inductive hypothesis applies to.



          For example, when proving Euler's formula for acyclic graphs, you could first argue (in your favorite way) that any such graph contains a vertex of degree $1$, then delete that vertex together with the edge out of it. This reduces the graph to a smaller one for which you have already verified Euler's formula.



          Here, there is no worry that we've gotten all of the graphs we're considering in this case, because we started with an arbitrary such graph.



          Aside from this, your strategy looks good. You should just rethink your arguments in terms of removing an edge or vertex, rather than adding them.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 53 mins ago









          Misha LavrovMisha Lavrov

          49.8k759109




          49.8k759109












          • $begingroup$
            Thanks! This is very informative.
            $endgroup$
            – TheLast Cipher
            37 mins ago


















          • $begingroup$
            Thanks! This is very informative.
            $endgroup$
            – TheLast Cipher
            37 mins ago
















          $begingroup$
          Thanks! This is very informative.
          $endgroup$
          – TheLast Cipher
          37 mins ago




          $begingroup$
          Thanks! This is very informative.
          $endgroup$
          – TheLast Cipher
          37 mins ago


















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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029