{
  "index": "2012-A-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $*$ be a commutative and associative binary operation on a set $S$. Assume that for every $x$\nand $y$ in $S$, there exists $z$ in $S$ such that $x * z = y$. (This $z$ may depend on $x$ and $y$.)\nShow that if $a,b,c$ are in $S$ and $a*c = b*c$, then $a=b$.",
  "solution": "Write $d$ for $a * c = b * c \\in S$. For some $e\\in S$, $d * e = a$, and\nthus for $f = c * e$, $a * f = a * c * e = d * e = a$ and $b * f = b * c\n* e = d * e = a$. Let $g \\in S$ satisfy $g * a = b$; then $b = g * a = g\n* (a * f) = (g * a) * f = b * f = a$, as desired.\n\n\\noindent\n\\textbf{Remark.} With slightly more work, one can show that $S$ forms an abelian group\nwith the operation $*$.",
  "vars": [
    "x",
    "y",
    "z",
    "a",
    "b",
    "c",
    "d",
    "e",
    "f",
    "g"
  ],
  "params": [
    "S"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "z": "variablez",
        "a": "elementa",
        "b": "elementb",
        "c": "elementc",
        "d": "elementd",
        "e": "elemente",
        "f": "elementf",
        "g": "elementg",
        "S": "setspace"
      },
      "question": "Let $*$ be a commutative and associative binary operation on a set $setspace$. Assume that for every $variablex$ and $variabley$ in $setspace$, there exists $variablez$ in $setspace$ such that $variablex * variablez = variabley$. (This $variablez$ may depend on $variablex$ and $variabley$.)\nShow that if $elementa,elementb,elementc$ are in $setspace$ and $elementa*elementc = elementb*elementc$, then $elementa=elementb$.",
      "solution": "Write $elementd$ for $elementa * elementc = elementb * elementc \\in setspace$. For some $elemente\\in setspace$, $elementd * elemente = elementa$, and\nthus for $elementf = elementc * elemente$, $elementa * elementf = elementa * elementc * elemente = elementd * elemente = elementa$ and $elementb * elementf = elementb * elementc\n* elemente = elementd * elemente = elementa$. Let $elementg \\in setspace$ satisfy $elementg * elementa = elementb$; then $elementb = elementg * elementa = elementg\n* (elementa * elementf) = (elementg * elementa) * elementf = elementb * elementf = elementa$, as desired.\n\n\\noindent\n\\textbf{Remark.} With slightly more work, one can show that $setspace$ forms an abelian group\nwith the operation $*$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "y": "buttermilk",
        "z": "peppermint",
        "a": "sunflower",
        "b": "lighthouse",
        "c": "raspberry",
        "d": "hummingbird",
        "e": "sailcloth",
        "f": "windflower",
        "g": "parchment",
        "S": "chandelier"
      },
      "question": "Let $*$ be a commutative and associative binary operation on a set $chandelier$. Assume that for every $sandstone$\nand $buttermilk$ in $chandelier$, there exists $peppermint$ in $chandelier$ such that $sandstone * peppermint = buttermilk$. (This $peppermint$ may depend on $sandstone$ and $buttermilk$.)\nShow that if $sunflower,lighthouse,raspberry$ are in $chandelier$ and $sunflower*raspberry = lighthouse*raspberry$, then $sunflower=lighthouse$.",
      "solution": "Write $hummingbird$ for $sunflower * raspberry = lighthouse * raspberry \\in chandelier$. For some $sailcloth\\in chandelier$, $hummingbird * sailcloth = sunflower$, and\nthus for $windflower = raspberry * sailcloth$, $sunflower * windflower = sunflower * raspberry * sailcloth = hummingbird * sailcloth = sunflower$ and $lighthouse * windflower = lighthouse * raspberry\n* sailcloth = hummingbird * sailcloth = sunflower$. Let $parchment \\in chandelier$ satisfy $parchment * sunflower = lighthouse$; then $lighthouse = parchment * sunflower = parchment\n* (sunflower * windflower) = (parchment * sunflower) * windflower = lighthouse * windflower = sunflower$, as desired.\n\n\\noindent\n\\textbf{Remark.} With slightly more work, one can show that $chandelier$ forms an abelian group\nwith the operation $*$.}"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "y": "settledvalue",
        "z": "destination",
        "a": "resultvalue",
        "b": "outputvalue",
        "c": "independent",
        "d": "originvalue",
        "e": "sourcevalue",
        "f": "mediator",
        "g": "reducer",
        "S": "singularity"
      },
      "question": "Let $*$ be a commutative and associative binary operation on a set $singularity$. Assume that for every $constantvalue$\nand $settledvalue$ in $singularity$, there exists $destination$ in $singularity$ such that $constantvalue * destination = settledvalue$. (This $destination$ may depend on $constantvalue$ and $settledvalue$.)\nShow that if $resultvalue,outputvalue,independent$ are in $singularity$ and $resultvalue*independent = outputvalue*independent$, then $resultvalue=outputvalue$.",
      "solution": "Write $originvalue$ for $resultvalue * independent = outputvalue * independent \\in singularity$. For some $sourcevalue\\in singularity$, $originvalue * sourcevalue = resultvalue$, and\nthus for $mediator = independent * sourcevalue$, $resultvalue * mediator = resultvalue * independent * sourcevalue = originvalue * sourcevalue = resultvalue$ and $outputvalue * mediator = outputvalue * independent\n* sourcevalue = originvalue * sourcevalue = resultvalue$. Let $reducer \\in singularity$ satisfy $reducer * resultvalue = outputvalue$; then $outputvalue = reducer * resultvalue = reducer\n* (resultvalue * mediator) = (reducer * resultvalue) * mediator = outputvalue * mediator = resultvalue$, as desired.\n\n\\noindent\n\\textbf{Remark.} With slightly more work, one can show that $singularity$ forms an abelian group\nwith the operation $*$.}",
      "error": false
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "z": "mnbvcxas",
        "a": "poiulkjh",
        "b": "lkjhgfds",
        "c": "qazwsxed",
        "d": "plmoknji",
        "e": "ijuhygtr",
        "f": "okmijnbh",
        "g": "nhytgbvf",
        "S": "werplmnb"
      },
      "question": "Let $*$ be a commutative and associative binary operation on a set $werplmnb$. Assume that for every $qzxwvtnp$\nand $hjgrksla$ in $werplmnb$, there exists $mnbvcxas$ in $werplmnb$ such that $qzxwvtnp * mnbvcxas = hjgrksla$. (This $mnbvcxas$ may depend on $qzxwvtnp$ and $hjgrksla$.)\nShow that if $poiulkjh,lkjhgfds,qazwsxed$ are in $werplmnb$ and $poiulkjh*qazwsxed = lkjhgfds*qazwsxed$, then $poiulkjh=lkjhgfds$.",
      "solution": "Write $plmoknji$ for $poiulkjh * qazwsxed = lkjhgfds * qazwsxed \\\\in werplmnb$. For some $ijuhygtr\\\\in werplmnb$, $plmoknji * ijuhygtr = poiulkjh$, and\nthus for $okmijnbh = qazwsxed * ijuhygtr$, $poiulkjh * okmijnbh = poiulkjh * qazwsxed * ijuhygtr = plmoknji * ijuhygtr = poiulkjh$ and $lkjhgfds * okmijnbh = lkjhgfds * qazwsxed\n* ijuhygtr = plmoknji * ijuhygtr = poiulkjh$. Let $nhytgbvf \\\\in werplmnb$ satisfy $nhytgbvf * poiulkjh = lkjhgfds$; then $lkjhgfds = nhytgbvf * poiulkjh = nhytgbvf\n* (poiulkjh * okmijnbh) = (nhytgbvf * poiulkjh) * okmijnbh = lkjhgfds * okmijnbh = poiulkjh$, as desired.\n\n\\\\noindent\n\\\\textbf{Remark.} With slightly more work, one can show that $werplmnb$ forms an abelian group\nwith the operation $*$.}   "
    },
    "kernel_variant": {
      "question": "Let $S$ be a non–empty set and let  \n\\[\n*\\;:\\;S\\times S\\longrightarrow S,\\qquad (x,y)\\longmapsto x*y\n\\]\nbe a binary operation satisfying  \n\n(1)\\; (associativity and commutativity)  \n\\[\n(x*y)*z \\;=\\; x*(y*z),\\qquad x*y \\;=\\; y*x\\quad(\\forall\\,x,y,z\\!\\in\\! S),\n\\]\n\n(2)\\; (right–division / solvability)  \n\\[\n\\forall\\,x,y\\in S\\;\\exists\\,z\\in S\\text{ such that }x*z=y .\n\\]\n\nFor every $c\\in S$ write  \n\\[\n\\lambda_{c}:S\\longrightarrow S,\\qquad \\lambda_{c}(x):=x*c\n\\]\n(“right multiplication by $c$’’).\n\n(a)\\; Prove that there exists a unique element $0\\in S$ such that  \n\\[\nx*0 = x\\qquad(\\forall\\,x\\in S).\n\\]\n\n(b)\\; Prove that every map $\\lambda_{c}$ is bijective and that for each\n$x\\in S$ there is a (necessarily unique) element $-x\\in S$ with  \n\\[\nx*(-x)=0 .\n\\]\nDeduce that $(S,*)$ is an abelian group with identity $0$ and inverse\nmap $x\\mapsto -x$.\n\n(c)\\; Let $\\Phi:S\\longrightarrow S$ be any bijection and define  \n\\[\nx\\star y := \\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(y)\\bigr)\\qquad(x,y\\in S).\n\\]\nShow that $(S,\\star)$ is again an abelian group, that its identity is\n\\[\ne_{\\star}= \\Phi^{-1}(0),\n\\]\nand that the right–division axiom (2) is still valid for $\\star$.\n\n(d)\\; Conversely, let $\\Phi:S\\to S$ be an arbitrary bijection and let\n$\\star$ be defined exactly as in {\\rm(c)}.  \nProve that $\\star$ satisfies {\\rm(1)} and {\\rm(2)}.  (Thus every binary\noperation obtained from $*$ by “transport of structure’’ along a\nbijection enjoys the same two axioms.)\n\n(e)\\; (Strong $k$–fold cancellation)  \nLet $k\\ge 1$ and let $a_{1},\\dots ,a_{k},\\;b_{1},\\dots ,b_{k},\\;c\\in S$\nsatisfy $a_{i}*c=b_{i}*c$ for $i=1,\\dots ,k$.  Show that  \n\\[\na_{1}*\\dots *a_{k}\\;=\\; b_{1}*\\dots *b_{k}.\n\\]\n\n--------------------------------------------------------------------",
      "solution": "Throughout, equalities are understood to hold for arbitrary elements of\n$S$ unless explicitly restricted.\n\n1.\\; Existence and uniqueness of an identity (part (a))  \nChoose an arbitrary $a\\in S$.  By (2) there exists $e\\in S$ with\n$a*e=a$.  Now fix $x\\in S$ and apply (2) once more to obtain\n$y\\in S$ such that $a*y=x$.  Then\n\\[\nx*e=(a*y)*e=a*(y*e)=(a*e)*y=a*y=x,\n\\]\nso $e$ is a right identity.  Because $*$ is commutative, $e$ is also a\nleft identity, hence an identity.  If $e'$ were another identity then\n$e=e*e'=e'$, so the identity is unique; denote it by $0$.\n\n2.\\; Bijectivity of the right translations and existence of inverses  \n(part (b))\n\nFix $c\\in S$.\n\nSurjectivity of $\\lambda_{c}$.  \nLet $y\\in S$.  By (2) with $x=c$ there exists $z\\in S$ such that\n$c*z=y$.  Using commutativity we have $z*c=y$, i.e.\\ $\\lambda_{c}(z)=y$.\nHence $\\lambda_{c}$ is surjective.\n\nInjectivity of $\\lambda_{c}$.  \nSuppose $\\lambda_{c}(x_{1})=\\lambda_{c}(x_{2})$, i.e.\\\n$x_{1}*c=x_{2}*c$.  \nBy (2) with $x=c,\\;y=0$ there exists $d\\in S$ such that $c*d=0$.\nAssociativity gives\n\\[\nx_{1} = x_{1}*0 = x_{1}*(c*d) = (x_{1}*c)*d\n       = (x_{2}*c)*d = x_{2}*(c*d) = x_{2}*0 = x_{2},\n\\]\nso $\\lambda_{c}$ is injective.  Consequently every $\\lambda_{c}$ is\nbijective.\n\nInverses.\nFor $a\\in S$, axiom (2) with $x=a,\\;y=0$ furnishes $u\\in S$ such that\n$a*u=0$.  Define $-a:=u$.  If $v$ is another element with $a*v=0$, then\n$\\lambda_{a}(u)=\\lambda_{a}(v)$, and injectivity of $\\lambda_{a}$\nforces $u=v$.  Thus $-a$ is uniquely determined.  With identity\n$0$ and inverse map $x\\mapsto -x$, the structure $(S,*)$ is an abelian\ngroup.\n\n3.\\; Transported operation is again a group (part (c))\n\nLet $\\Phi$ be a bijection and define\n\\[\nx\\star y:=\\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(y)\\bigr).\n\\]\nThe map  \n\\[\n\\Phi:(S,\\star)\\longrightarrow(S,*),\\qquad x\\longmapsto\\Phi(x)\n\\]\nis a bijective homomorphism, since\n\\[\n\\Phi(x\\star y)=\\Phi\\!\\bigl(\\Phi^{-1}(\\Phi(x)*\\Phi(y))\\bigr)\n             =\\Phi(x)*\\Phi(y).\n\\]\nHence $(S,\\star)$ is an abelian group.  Its identity $e_{\\star}$ is the\nunique element satisfying $\\Phi(e_{\\star})=0$, namely\n$e_{\\star}=\\Phi^{-1}(0)$.\n\nRight–division for $\\star$.  \nFix $x,y\\in S$.  Because every $\\lambda_{c}$ is bijective in $(S,*)$,\nthe map\n\\[\nz\\longmapsto x\\star z\n          =\\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(z)\\bigr)\n          =\\Phi^{-1}\\!\\bigl(\\lambda_{\\Phi(x)}(\\Phi(z))\\bigr)\n\\]\nis a composition of bijections, hence bijective.  Therefore for every\n$y$ there exists $z$ with $x\\star z=y$, so axiom (2) holds for $\\star$.\n\n4.\\; Every transported operation enjoys (1) and (2) (part (d))\n\nConversely, start with a bijection $\\Phi$ and define\n\\[\nx\\star y=\\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(y)\\bigr).\n\\]\nAssociativity and commutativity of $\\star$ follow immediately from the\ncorresponding properties of $*$, because $\\Phi$ is a bijection.  The\nargument for right–division given in part (c) does not use any special\nproperty of $\\Phi$ beyond bijectivity, hence applies verbatim; thus\n$\\star$ satisfies (1) and (2).\n\n5.\\; Strong $k$–fold cancellation (part (e))\n\nBecause $(S,*)$ is an abelian group, for $i=1,\\dots ,k$ the hypothesis\n$a_{i}*c=b_{i}*c$ implies\n\\[\na_{i}*c*(c^{-1}) = b_{i}*c*(c^{-1})\\;\\Longrightarrow\\;a_{i}=b_{i},\n\\]\nwhere $c^{-1}=-c$ is the group inverse of $c$.  Multiplying the $k$\nequalities together gives\n\\[\na_{1}*\\dots *a_{k}=b_{1}*\\dots *b_{k},\n\\]\nas required.\n\nAll parts are thus rigorously proved.  In particular, the two axioms\n(1) and (2) already force $(S,*)$ to be an abelian group; any binary\noperation obtained by transporting $*$ along a bijection again satisfies\nthese axioms; and the $k$–fold cancellation law follows immediately\nfrom the group structure.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.822603",
        "was_fixed": false,
        "difficulty_analysis": "1. Two operations instead of one.  The problem no longer concerns a single binary law but asks the solver to control a pair of associative–commutative laws that interact through non-trivial “exchange” identities (E1,E2).\n\n2. Deeper structural goal.  One must first recognise, then prove, that each law separately makes S into an abelian group; this already subsumes the entire original problem (identity + inverses + cancellation).  Only after that does the real difficulty—showing the two laws coincide—begin.\n\n3. Non-obvious use of group automorphisms.  A key insight is to package the interaction (E1) as an automorphism φ of the ★-group and then analyse that automorphism (show φ²=id ⇒ φ=id).  This is far beyond the simple cancellation argument of the kernel variant.\n\n4. Multiple interacting concepts.  The solver must simultaneously handle right-division, cancellation, identities, inverses, automorphisms of abelian groups, and a novel system of functional equations (exchange laws).\n\n5. Stronger conclusion.  After building the abelian-group structure, the problem culminates in a multi-term cancellation principle (part (c)), which requires combining group theory with careful bookkeeping of products and exponents.\n\nAltogether these additions force the contestant to organise a substantially longer argument that draws on higher-level algebraic notions, far exceeding the scope and technical depth of the original kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let $S$ be a non–empty set and let  \n\\[\n*\\;:\\;S\\times S\\longrightarrow S,\\qquad (x,y)\\longmapsto x*y\n\\]\nbe a binary operation satisfying  \n\n(1)\\; (associativity and commutativity)  \n\\[\n(x*y)*z \\;=\\; x*(y*z),\\qquad x*y \\;=\\; y*x\\quad(\\forall\\,x,y,z\\!\\in\\! S),\n\\]\n\n(2)\\; (right–division / solvability)  \n\\[\n\\forall\\,x,y\\in S\\;\\exists\\,z\\in S\\text{ such that }x*z=y .\n\\]\n\nFor every $c\\in S$ write  \n\\[\n\\lambda_{c}:S\\longrightarrow S,\\qquad \\lambda_{c}(x):=x*c\n\\]\n(“right multiplication by $c$’’).\n\n(a)\\; Prove that there exists a unique element $0\\in S$ such that  \n\\[\nx*0 = x\\qquad(\\forall\\,x\\in S).\n\\]\n\n(b)\\; Prove that every map $\\lambda_{c}$ is bijective and that for each\n$x\\in S$ there is a (necessarily unique) element $-x\\in S$ with  \n\\[\nx*(-x)=0 .\n\\]\nDeduce that $(S,*)$ is an abelian group with identity $0$ and inverse\nmap $x\\mapsto -x$.\n\n(c)\\; Let $\\Phi:S\\longrightarrow S$ be any bijection and define  \n\\[\nx\\star y := \\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(y)\\bigr)\\qquad(x,y\\in S).\n\\]\nShow that $(S,\\star)$ is again an abelian group, that its identity is\n\\[\ne_{\\star}= \\Phi^{-1}(0),\n\\]\nand that the right–division axiom (2) is still valid for $\\star$.\n\n(d)\\; Conversely, let $\\Phi:S\\to S$ be an arbitrary bijection and let\n$\\star$ be defined exactly as in {\\rm(c)}.  \nProve that $\\star$ satisfies {\\rm(1)} and {\\rm(2)}.  (Thus every binary\noperation obtained from $*$ by “transport of structure’’ along a\nbijection enjoys the same two axioms.)\n\n(e)\\; (Strong $k$–fold cancellation)  \nLet $k\\ge 1$ and let $a_{1},\\dots ,a_{k},\\;b_{1},\\dots ,b_{k},\\;c\\in S$\nsatisfy $a_{i}*c=b_{i}*c$ for $i=1,\\dots ,k$.  Show that  \n\\[\na_{1}*\\dots *a_{k}\\;=\\; b_{1}*\\dots *b_{k}.\n\\]\n\n--------------------------------------------------------------------",
      "solution": "Throughout, equalities are understood to hold for arbitrary elements of\n$S$ unless explicitly restricted.\n\n1.\\; Existence and uniqueness of an identity (part (a))  \nChoose an arbitrary $a\\in S$.  By (2) there exists $e\\in S$ with\n$a*e=a$.  Now fix $x\\in S$ and apply (2) once more to obtain\n$y\\in S$ such that $a*y=x$.  Then\n\\[\nx*e=(a*y)*e=a*(y*e)=(a*e)*y=a*y=x,\n\\]\nso $e$ is a right identity.  Because $*$ is commutative, $e$ is also a\nleft identity, hence an identity.  If $e'$ were another identity then\n$e=e*e'=e'$, so the identity is unique; denote it by $0$.\n\n2.\\; Bijectivity of the right translations and existence of inverses  \n(part (b))\n\nFix $c\\in S$.\n\nSurjectivity of $\\lambda_{c}$.  \nLet $y\\in S$.  By (2) with $x=c$ there exists $z\\in S$ such that\n$c*z=y$.  Using commutativity we have $z*c=y$, i.e.\\ $\\lambda_{c}(z)=y$.\nHence $\\lambda_{c}$ is surjective.\n\nInjectivity of $\\lambda_{c}$.  \nSuppose $\\lambda_{c}(x_{1})=\\lambda_{c}(x_{2})$, i.e.\\\n$x_{1}*c=x_{2}*c$.  \nBy (2) with $x=c,\\;y=0$ there exists $d\\in S$ such that $c*d=0$.\nAssociativity gives\n\\[\nx_{1} = x_{1}*0 = x_{1}*(c*d) = (x_{1}*c)*d\n       = (x_{2}*c)*d = x_{2}*(c*d) = x_{2}*0 = x_{2},\n\\]\nso $\\lambda_{c}$ is injective.  Consequently every $\\lambda_{c}$ is\nbijective.\n\nInverses.\nFor $a\\in S$, axiom (2) with $x=a,\\;y=0$ furnishes $u\\in S$ such that\n$a*u=0$.  Define $-a:=u$.  If $v$ is another element with $a*v=0$, then\n$\\lambda_{a}(u)=\\lambda_{a}(v)$, and injectivity of $\\lambda_{a}$\nforces $u=v$.  Thus $-a$ is uniquely determined.  With identity\n$0$ and inverse map $x\\mapsto -x$, the structure $(S,*)$ is an abelian\ngroup.\n\n3.\\; Transported operation is again a group (part (c))\n\nLet $\\Phi$ be a bijection and define\n\\[\nx\\star y:=\\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(y)\\bigr).\n\\]\nThe map  \n\\[\n\\Phi:(S,\\star)\\longrightarrow(S,*),\\qquad x\\longmapsto\\Phi(x)\n\\]\nis a bijective homomorphism, since\n\\[\n\\Phi(x\\star y)=\\Phi\\!\\bigl(\\Phi^{-1}(\\Phi(x)*\\Phi(y))\\bigr)\n             =\\Phi(x)*\\Phi(y).\n\\]\nHence $(S,\\star)$ is an abelian group.  Its identity $e_{\\star}$ is the\nunique element satisfying $\\Phi(e_{\\star})=0$, namely\n$e_{\\star}=\\Phi^{-1}(0)$.\n\nRight–division for $\\star$.  \nFix $x,y\\in S$.  Because every $\\lambda_{c}$ is bijective in $(S,*)$,\nthe map\n\\[\nz\\longmapsto x\\star z\n          =\\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(z)\\bigr)\n          =\\Phi^{-1}\\!\\bigl(\\lambda_{\\Phi(x)}(\\Phi(z))\\bigr)\n\\]\nis a composition of bijections, hence bijective.  Therefore for every\n$y$ there exists $z$ with $x\\star z=y$, so axiom (2) holds for $\\star$.\n\n4.\\; Every transported operation enjoys (1) and (2) (part (d))\n\nConversely, start with a bijection $\\Phi$ and define\n\\[\nx\\star y=\\Phi^{-1}\\!\\bigl(\\Phi(x)*\\Phi(y)\\bigr).\n\\]\nAssociativity and commutativity of $\\star$ follow immediately from the\ncorresponding properties of $*$, because $\\Phi$ is a bijection.  The\nargument for right–division given in part (c) does not use any special\nproperty of $\\Phi$ beyond bijectivity, hence applies verbatim; thus\n$\\star$ satisfies (1) and (2).\n\n5.\\; Strong $k$–fold cancellation (part (e))\n\nBecause $(S,*)$ is an abelian group, for $i=1,\\dots ,k$ the hypothesis\n$a_{i}*c=b_{i}*c$ implies\n\\[\na_{i}*c*(c^{-1}) = b_{i}*c*(c^{-1})\\;\\Longrightarrow\\;a_{i}=b_{i},\n\\]\nwhere $c^{-1}=-c$ is the group inverse of $c$.  Multiplying the $k$\nequalities together gives\n\\[\na_{1}*\\dots *a_{k}=b_{1}*\\dots *b_{k},\n\\]\nas required.\n\nAll parts are thus rigorously proved.  In particular, the two axioms\n(1) and (2) already force $(S,*)$ to be an abelian group; any binary\noperation obtained by transporting $*$ along a bijection again satisfies\nthese axioms; and the $k$–fold cancellation law follows immediately\nfrom the group structure.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.629352",
        "was_fixed": false,
        "difficulty_analysis": "1. Two operations instead of one.  The problem no longer concerns a single binary law but asks the solver to control a pair of associative–commutative laws that interact through non-trivial “exchange” identities (E1,E2).\n\n2. Deeper structural goal.  One must first recognise, then prove, that each law separately makes S into an abelian group; this already subsumes the entire original problem (identity + inverses + cancellation).  Only after that does the real difficulty—showing the two laws coincide—begin.\n\n3. Non-obvious use of group automorphisms.  A key insight is to package the interaction (E1) as an automorphism φ of the ★-group and then analyse that automorphism (show φ²=id ⇒ φ=id).  This is far beyond the simple cancellation argument of the kernel variant.\n\n4. Multiple interacting concepts.  The solver must simultaneously handle right-division, cancellation, identities, inverses, automorphisms of abelian groups, and a novel system of functional equations (exchange laws).\n\n5. Stronger conclusion.  After building the abelian-group structure, the problem culminates in a multi-term cancellation principle (part (c)), which requires combining group theory with careful bookkeeping of products and exponents.\n\nAltogether these additions force the contestant to organise a substantially longer argument that draws on higher-level algebraic notions, far exceeding the scope and technical depth of the original kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}