{
  "index": "1956-B-2",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. Suppose that each set \\( X \\) of points in the plane has an associated set \\( \\bar{X} \\) of points called its cover. Suppose further that\n(1) \\( \\bar{X} \\cup Y \\overline{\\bar{X}} \\cup \\bar{Y} \\cup Y \\), where \\( \\cup \\) designates point set sum (or union) and \\( \\supset \\) denotes set inclusion.\nPtove: (i) \\( \\bar{X} \\supset X \\), (ii) \\( \\overline{\\bar{X}}=\\bar{X} \\), (iii) \\( X \\supset Y \\) implies \\( \\bar{X} \\supset \\bar{Y} \\).\nProve conversely that (i), (ii) and (iii) imply (1).",
  "solution": "Solution. In (1) let \\( Y=X \\) : then\n\\[\n\\bar{X} \\supset \\overline{\\bar{X}} \\cup \\bar{X} \\cup X\n\\]\nfrom which it is clear that \\( \\bar{X} \\supset X \\), which is (i).\nWe note from the above that \\( \\bar{X} \\supset \\overline{\\bar{X}} \\). In (i) we replace \\( X \\) by \\( \\bar{X} \\) to get \\( \\overline{\\bar{X}} \\supset \\bar{X} \\). These two relations imply that \\( \\overline{\\bar{X}}=\\bar{X} \\), which is (ii).\n\nSuppose \\( Y \\subset X \\). Then \\( X \\cup Y=X \\), and (1) reduces to\n\\[\n\\bar{X} \\supset \\overline{\\bar{X}} \\cup \\bar{Y} \\cup Y\n\\]\nwhence \\( \\bar{X} \\supset \\bar{Y} \\), which is (iii).\nNow suppose that (i), (ii), and (iii) hold. For any sets \\( X \\) and \\( Y \\) we have\n\\[\nX \\subset X \\cup Y .\n\\]\n\nBy (iii),\n\\[\n\\bar{X} \\subset \\bar{X} \\cup \\bar{Y} .\n\\]\n\nHence by (ii)\n\\[\n\\overline{\\bar{X}} \\subset \\bar{X} \\cup \\bar{Y} .\n\\]\n\nAlso\n\\[\nY \\subset X \\cup Y\n\\]\nand by (iii)\n\\[\n\\bar{Y} \\subset \\bar{X} \\cup Y .\n\\]\n\nFrom (i)\n\\[\nX \\cup Y \\subset \\bar{X} \\cup \\bar{Y}\n\\]\nso\n\\[\nY \\subset \\overline{X \\cup Y} .\n\\]\n\nNow (2), (3) and (4) together yield\n\\[\n\\overline{X \\cup Y} \\supset \\overline{\\bar{X}} \\cup \\bar{Y} \\cup Y,\n\\]\nwhich is the required condition (1).\nRemark. See Garrett Birkhoff, Lattice Theory, Amer. Math. Soc. Colloquium Pub., vol. 25, Providence, R.I., 1967, page 113.",
  "vars": [
    "X",
    "Y"
  ],
  "params": [],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "X": "firstset",
        "Y": "secondset"
      },
      "question": "2. Suppose that each set \\( firstset \\) of points in the plane has an associated set \\( \\bar{firstset} \\) of points called its cover. Suppose further that\n(1) \\( \\bar{firstset} \\cup secondset \\overline{\\bar{firstset}} \\cup \\bar{secondset} \\cup secondset \\), where \\( \\cup \\) designates point set sum (or union) and \\( \\supset \\) denotes set inclusion.\nPtove: (i) \\( \\bar{firstset} \\supset firstset \\), (ii) \\( \\overline{\\bar{firstset}}=\\bar{firstset} \\), (iii) \\( firstset \\supset secondset \\) implies \\( \\bar{firstset} \\supset \\bar{secondset} \\).\nProve conversely that (i), (ii) and (iii) imply (1).",
      "solution": "Solution. In (1) let \\( secondset=firstset \\) : then\n\\[\n\\bar{firstset} \\supset \\overline{\\bar{firstset}} \\cup \\bar{firstset} \\cup firstset\n\\]\nfrom which it is clear that \\( \\bar{firstset} \\supset firstset \\), which is (i).\nWe note from the above that \\( \\bar{firstset} \\supset \\overline{\\bar{firstset}} \\). In (i) we replace \\( firstset \\) by \\( \\bar{firstset} \\) to get \\( \\overline{\\bar{firstset}} \\supset \\bar{firstset} \\). These two relations imply that \\( \\overline{\\bar{firstset}}=\\bar{firstset} \\), which is (ii).\n\nSuppose \\( secondset \\subset firstset \\). Then \\( firstset \\cup secondset=firstset \\), and (1) reduces to\n\\[\n\\bar{firstset} \\supset \\overline{\\bar{firstset}} \\cup \\bar{secondset} \\cup secondset\n\\]\nwhence \\( \\bar{firstset} \\supset \\bar{secondset} \\), which is (iii).\nNow suppose that (i), (ii), and (iii) hold. For any sets \\( firstset \\) and \\( secondset \\) we have\n\\[\nfirstset \\subset firstset \\cup secondset .\n\\]\n\nBy (iii),\n\\[\n\\bar{firstset} \\subset \\bar{firstset} \\cup \\bar{secondset} .\n\\]\n\nHence by (ii)\n\\[\n\\overline{\\bar{firstset}} \\subset \\bar{firstset} \\cup \\bar{secondset} .\n\\]\n\nAlso\n\\[\nsecondset \\subset firstset \\cup secondset\n\\]\nand by (iii)\n\\[\n\\bar{secondset} \\subset \\bar{firstset} \\cup secondset .\n\\]\n\nFrom (i)\n\\[\nfirstset \\cup secondset \\subset \\bar{firstset} \\cup \\bar{secondset}\n\\]\nso\n\\[\nsecondset \\subset \\overline{firstset \\cup secondset} .\n\\]\n\nNow (2), (3) and (4) together yield\n\\[\n\\overline{firstset \\cup secondset} \\supset \\overline{\\bar{firstset}} \\cup \\bar{secondset} \\cup secondset,\n\\]\nwhich is the required condition (1).\nRemark. See Garrett Birkhoff, Lattice Theory, Amer. Math. Soc. Colloquium Pub., vol. 25, Providence, R.I., 1967, page 113."
    },
    "descriptive_long_confusing": {
      "map": {
        "X": "perimeter",
        "Y": "lighthouse"
      },
      "question": "2. Suppose that each set \\( perimeter \\) of points in the plane has an associated set \\( \\bar{perimeter} \\) of points called its cover. Suppose further that\n(1) \\( \\bar{perimeter} \\cup lighthouse \\overline{\\bar{perimeter}} \\cup \\bar{lighthouse} \\cup lighthouse \\), where \\( \\cup \\) designates point set sum (or union) and \\( \\supset \\) denotes set inclusion.\nPtove: (i) \\( \\bar{perimeter} \\supset perimeter \\), (ii) \\( \\overline{\\bar{perimeter}}=\\bar{perimeter} \\), (iii) \\( perimeter \\supset lighthouse \\) implies \\( \\bar{perimeter} \\supset \\bar{lighthouse} \\).\nProve conversely that (i), (ii) and (iii) imply (1).",
      "solution": "Solution. In (1) let \\( lighthouse=perimeter \\) : then\n\\[\n\\bar{perimeter} \\supset \\overline{\\bar{perimeter}} \\cup \\bar{perimeter} \\cup perimeter\n\\]\nfrom which it is clear that \\( \\bar{perimeter} \\supset perimeter \\), which is (i).\nWe note from the above that \\( \\bar{perimeter} \\supset \\overline{\\bar{perimeter}} \\). In (i) we replace \\( perimeter \\) by \\( \\bar{perimeter} \\) to get \\( \\overline{\\bar{perimeter}} \\supset \\bar{perimeter} \\). These two relations imply that \\( \\overline{\\bar{perimeter}}=\\bar{perimeter} \\), which is (ii).\n\nSuppose \\( lighthouse \\subset perimeter \\). Then \\( perimeter \\cup lighthouse=perimeter \\), and (1) reduces to\n\\[\n\\bar{perimeter} \\supset \\overline{\\bar{perimeter}} \\cup \\bar{lighthouse} \\cup lighthouse\n\\]\nwhence \\( \\bar{perimeter} \\supset \\bar{lighthouse} \\), which is (iii).\nNow suppose that (i), (ii), and (iii) hold. For any sets \\( perimeter \\) and \\( lighthouse \\) we have\n\\[\nperimeter \\subset perimeter \\cup lighthouse .\n\\]\n\nBy (iii),\n\\[\n\\bar{perimeter} \\subset \\bar{perimeter} \\cup \\bar{lighthouse} .\n\\]\n\nHence by (ii)\n\\[\n\\overline{\\bar{perimeter}} \\subset \\bar{perimeter} \\cup \\bar{lighthouse} .\n\\]\n\nAlso\n\\[\nlighthouse \\subset perimeter \\cup lighthouse\n\\]\nand by (iii)\n\\[\n\\bar{lighthouse} \\subset \\bar{perimeter} \\cup lighthouse .\n\\]\n\nFrom (i)\n\\[\nperimeter \\cup lighthouse \\subset \\bar{perimeter} \\cup \\bar{lighthouse}\n\\]\nso\n\\[\nlighthouse \\subset \\overline{perimeter \\cup lighthouse} .\n\\]\n\nNow (2), (3) and (4) together yield\n\\[\n\\overline{perimeter \\cup lighthouse} \\supset \\overline{\\bar{perimeter}} \\cup \\bar{lighthouse} \\cup lighthouse,\n\\]\nwhich is the required condition (1).\nRemark. See Garrett Birkhoff, Lattice Theory, Amer. Math. Soc. Colloquium Pub., vol. 25, Providence, R.I., 1967, page 113."
    },
    "descriptive_long_misleading": {
      "map": {
        "X": "knownset",
        "Y": "fixedset"
      },
      "question": "2. Suppose that each set \\( knownset \\) of points in the plane has an associated set \\( \\bar{knownset} \\) of points called its cover. Suppose further that\n(1) \\( \\bar{knownset} \\cup fixedset \\overline{\\bar{knownset}} \\cup \\bar{fixedset} \\cup fixedset \\), where \\( \\cup \\) designates point set sum (or union) and \\( \\supset \\) denotes set inclusion.\nPtove: (i) \\( \\bar{knownset} \\supset knownset \\), (ii) \\( \\overline{\\bar{knownset}}=\\bar{knownset} \\), (iii) \\( knownset \\supset fixedset \\) implies \\( \\bar{knownset} \\supset \\bar{fixedset} \\).\nProve conversely that (i), (ii) and (iii) imply (1).",
      "solution": "Solution. In (1) let \\( fixedset=knownset \\) : then\n\\[\n\\bar{knownset} \\supset \\overline{\\bar{knownset}} \\cup \\bar{knownset} \\cup knownset\n\\]\nfrom which it is clear that \\( \\bar{knownset} \\supset knownset \\), which is (i).\nWe note from the above that \\( \\bar{knownset} \\supset \\overline{\\bar{knownset}} \\). In (i) we replace \\( knownset \\) by \\( \\bar{knownset} \\) to get \\( \\overline{\\bar{knownset}} \\supset \\bar{knownset} \\). These two relations imply that \\( \\overline{\\bar{knownset}}=\\bar{knownset} \\), which is (ii).\n\nSuppose \\( fixedset \\subset knownset \\). Then \\( knownset \\cup fixedset=knownset \\), and (1) reduces to\n\\[\n\\bar{knownset} \\supset \\overline{\\bar{knownset}} \\cup \\bar{fixedset} \\cup fixedset\n\\]\nwhence \\( \\bar{knownset} \\supset \\bar{fixedset} \\), which is (iii).\nNow suppose that (i), (ii), and (iii) hold. For any sets \\( knownset \\) and \\( fixedset \\) we have\n\\[\nknownset \\subset knownset \\cup fixedset .\n\\]\n\nBy (iii),\n\\[\n\\bar{knownset} \\subset \\bar{knownset} \\cup \\bar{fixedset} .\n\\]\n\nHence by (ii)\n\\[\n\\overline{\\bar{knownset}} \\subset \\bar{knownset} \\cup \\bar{fixedset} .\n\\]\n\nAlso\n\\[\nfixedset \\subset knownset \\cup fixedset\n\\]\nand by (iii)\n\\[\n\\bar{fixedset} \\subset \\bar{knownset} \\cup fixedset .\n\\]\n\nFrom (i)\n\\[\nknownset \\cup fixedset \\subset \\bar{knownset} \\cup \\bar{fixedset}\n\\]\nso\n\\[\nfixedset \\subset \\overline{knownset \\cup fixedset} .\n\\]\n\nNow (2), (3) and (4) together yield\n\\[\n\\overline{knownset \\cup fixedset} \\supset \\overline{\\bar{knownset}} \\cup \\bar{fixedset} \\cup fixedset,\n\\]\nwhich is the required condition (1).\nRemark. See Garrett Birkhoff, Lattice Theory, Amer. Math. Soc. Colloquium Pub., vol. 25, Providence, R.I., 1967, page 113."
    },
    "garbled_string": {
      "map": {
        "X": "qzxwvtnp",
        "Y": "hjgrksla"
      },
      "question": "2. Suppose that each set \\( qzxwvtnp \\) of points in the plane has an associated set \\( \\bar{qzxwvtnp} \\) of points called its cover. Suppose further that\n(1) \\( \\bar{qzxwvtnp} \\cup hjgrksla \\overline{\\bar{qzxwvtnp}} \\cup \\bar{hjgrksla} \\cup hjgrksla \\), where \\( \\cup \\) designates point set sum (or union) and \\( \\supset \\) denotes set inclusion.\nPtove: (i) \\( \\bar{qzxwvtnp} \\supset qzxwvtnp \\), (ii) \\( \\overline{\\bar{qzxwvtnp}}=\\bar{qzxwvtnp} \\), (iii) \\( qzxwvtnp \\supset hjgrksla \\) implies \\( \\bar{qzxwvtnp} \\supset \\bar{hjgrksla} \\).\nProve conversely that (i), (ii) and (iii) imply (1).",
      "solution": "Solution. In (1) let \\( hjgrksla=qzxwvtnp \\) : then\n\\[\n\\bar{qzxwvtnp} \\supset \\overline{\\bar{qzxwvtnp}} \\cup \\bar{qzxwvtnp} \\cup qzxwvtnp\n\\]\nfrom which it is clear that \\( \\bar{qzxwvtnp} \\supset qzxwvtnp \\), which is (i).\nWe note from the above that \\( \\bar{qzxwvtnp} \\supset \\overline{\\bar{qzxwvtnp}} \\). In (i) we replace \\( qzxwvtnp \\) by \\( \\bar{qzxwvtnp} \\) to get \\( \\overline{\\bar{qzxwvtnp}} \\supset \\bar{qzxwvtnp} \\). These two relations imply that \\( \\overline{\\bar{qzxwvtnp}}=\\bar{qzxwvtnp} \\), which is (ii).\n\nSuppose \\( hjgrksla \\subset qzxwvtnp \\). Then \\( qzxwvtnp \\cup hjgrksla=qzxwvtnp \\), and (1) reduces to\n\\[\n\\bar{qzxwvtnp} \\supset \\overline{\\bar{qzxwvtnp}} \\cup \\bar{hjgrksla} \\cup hjgrksla\n\\]\nwhence \\( \\bar{qzxwvtnp} \\supset \\bar{hjgrksla} \\), which is (iii).\nNow suppose that (i), (ii), and (iii) hold. For any sets \\( qzxwvtnp \\) and \\( hjgrksla \\) we have\n\\[\nqzxwvtnp \\subset qzxwvtnp \\cup hjgrksla .\n\\]\n\nBy (iii),\n\\[\n\\bar{qzxwvtnp} \\subset \\bar{qzxwvtnp} \\cup \\bar{hjgrksla} .\n\\]\n\nHence by (ii)\n\\[\n\\overline{\\bar{qzxwvtnp}} \\subset \\bar{qzxwvtnp} \\cup \\bar{hjgrksla} .\n\\]\n\nAlso\n\\[\nhjgrksla \\subset qzxwvtnp \\cup hjgrksla\n\\]\nand by (iii)\n\\[\n\\bar{hjgrksla} \\subset \\bar{qzxwvtnp} \\cup hjgrksla .\n\\]\n\nFrom (i)\n\\[\nqzxwvtnp \\cup hjgrksla \\subset \\bar{qzxwvtnp} \\cup \\bar{hjgrksla}\n\\]\nso\n\\[\nhjgrksla \\subset \\overline{qzxwvtnp \\cup hjgrksla} .\n\\]\n\nNow (2), (3) and (4) together yield\n\\[\n\\overline{qzxwvtnp \\cup hjgrksla} \\supset \\overline{\\bar{qzxwvtnp}} \\cup \\bar{hjgrksla} \\cup hjgrksla,\n\\]\nwhich is the required condition (1).\nRemark. See Garrett Birkhoff, Lattice Theory, Amer. Math. Soc. Colloquium Pub., vol. 25, Providence, R.I., 1967, page 113."
    },
    "kernel_variant": {
      "question": "Let S be a non–empty set.  \nTo every subset A ⊆ S assign another subset, denoted A★ and called the cloak of A.  \nAssume that for every triple of subsets A, B, C ⊆ S the following single axiom (the “triple–cloak inclusion”) holds:\n\n(1)  (A★)★ ∪ B★ ∪ C ⊆ (A ∪ B ∪ C)★.\n\nProve that the cloak operation satisfies  \n(i) Inflation    A ⊆ A★;  \n(ii) Idempotence  (A★)★ = A★;  \n(iii) Monotonicity  A ⊆ B ⇒ A★ ⊆ B★.  \n\nConversely, show that conditions (i)–(iii) alone force the triple–cloak inclusion (1).",
      "solution": "Forward direction: (1) ⇒ (i)–(iii).  \n------------------------------------------------\n(i) Inflation.  \nPut B = C = A in (1):\n\n (A★)★ ∪ A★ ∪ A ⊆ (A ∪ A ∪ A)★ = A★.\n\nBecause the right–hand side contains the whole left–hand side, in particular it contains A; hence A ⊆ A★.\n\n(ii) Idempotence.  \nWith B = C = A again we obtained\n\n (A★)★ ∪ A★ ∪ A ⊆ A★.  (2)\n\nThe right–hand side obviously contains A★; therefore (2) gives (A★)★ ⊆ A★.  \nOn the other hand, plugging A ← A★ (and keeping B = C = A★) into (1) gives\n\n ((A★)★)★ ∪ (A★)★ ∪ A★ ⊆ (A★)★,\n\nwhence A★ ⊆ (A★)★.  \nCombining the two inclusions yields (A★)★ = A★.\n\n(iii) Monotonicity.  \nAssume A ⊆ B.  Put B (in axiom (1)) equal to A and C equal to B:\n\n (A★)★ ∪ A★ ∪ B ⊆ (A ∪ A ∪ B)★ = B★.\n\nBy (ii) we have (A★)★ = A★, so the left–hand side is simply A★ ∪ B, which in particular contains A★.  Hence A★ ⊆ B★, proving monotonicity.\n\nBackward direction: (i)–(iii) ⇒ (1).  \n--------------------------------------\nLet A, B, C be arbitrary.  We have three easy containments.\n\n1.  A ⊆ A ∪ B ∪ C, so by (iii) A★ ⊆ (A ∪ B ∪ C)★.  \n With (ii) this yields (A★)★ = A★ ⊆ (A ∪ B ∪ C)★.\n\n2.  B ⊆ A ∪ B ∪ C, so again by (iii) B★ ⊆ (A ∪ B ∪ C)★.\n\n3.  Trivially C ⊆ A ∪ B ∪ C, hence by (i)  C ⊆ (A ∪ B ∪ C)★.\n\nPutting the three inclusions together gives\n\n (A★)★ ∪ B★ ∪ C ⊆ (A ∪ B ∪ C)★,\n\nwhich is exactly axiom (1).\n\nTherefore (1) and (i)–(iii) are equivalent, and the cloak operation enjoys inflation, idempotence and monotonicity as required. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.483218",
        "was_fixed": false,
        "difficulty_analysis": "• Several independent sets (A, B, C) now interact in a single axiom instead of just two; tracking how each of the three pieces behaves under the cloak map forces considerably more bookkeeping.\n\n• Both the cloak of a set (B★) and an uncloaked set (C) appear on the left of the axiom, so the proof must handle the interaction between “closed” and “raw” data simultaneously.\n\n• Deriving idempotence is subtler: it comes from recognising that the same axiom, specialised with B = C = A, simultaneously furnishes two opposite inclusions.  In the original kernel this step required only two sets; here one must keep three coordinated.\n\n• Monotonicity can no longer be read off directly from a single substitution.  One has to choose B and C cunningly (one cloaked, one uncloaked) so that A★ shows up explicitly inside the big union given by the axiom.\n\n• The converse direction likewise demands a triple-layer argument, feeding each of A★, B★ and C into properties (i)–(iii) separately before recombining them.\n\n• Overall the problem contains more variables, more cases to analyse, and a less obvious route to idempotence and monotonicity, making the reasoning chain distinctly longer and more intricate than in either the original problem or the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let S be a non–empty set.  \nTo every subset A ⊆ S assign another subset, denoted A★ and called the cloak of A.  \nAssume that for every triple of subsets A, B, C ⊆ S the following single axiom (the “triple–cloak inclusion”) holds:\n\n(1)  (A★)★ ∪ B★ ∪ C ⊆ (A ∪ B ∪ C)★.\n\nProve that the cloak operation satisfies  \n(i) Inflation    A ⊆ A★;  \n(ii) Idempotence  (A★)★ = A★;  \n(iii) Monotonicity  A ⊆ B ⇒ A★ ⊆ B★.  \n\nConversely, show that conditions (i)–(iii) alone force the triple–cloak inclusion (1).",
      "solution": "Forward direction: (1) ⇒ (i)–(iii).  \n------------------------------------------------\n(i) Inflation.  \nPut B = C = A in (1):\n\n (A★)★ ∪ A★ ∪ A ⊆ (A ∪ A ∪ A)★ = A★.\n\nBecause the right–hand side contains the whole left–hand side, in particular it contains A; hence A ⊆ A★.\n\n(ii) Idempotence.  \nWith B = C = A again we obtained\n\n (A★)★ ∪ A★ ∪ A ⊆ A★.  (2)\n\nThe right–hand side obviously contains A★; therefore (2) gives (A★)★ ⊆ A★.  \nOn the other hand, plugging A ← A★ (and keeping B = C = A★) into (1) gives\n\n ((A★)★)★ ∪ (A★)★ ∪ A★ ⊆ (A★)★,\n\nwhence A★ ⊆ (A★)★.  \nCombining the two inclusions yields (A★)★ = A★.\n\n(iii) Monotonicity.  \nAssume A ⊆ B.  Put B (in axiom (1)) equal to A and C equal to B:\n\n (A★)★ ∪ A★ ∪ B ⊆ (A ∪ A ∪ B)★ = B★.\n\nBy (ii) we have (A★)★ = A★, so the left–hand side is simply A★ ∪ B, which in particular contains A★.  Hence A★ ⊆ B★, proving monotonicity.\n\nBackward direction: (i)–(iii) ⇒ (1).  \n--------------------------------------\nLet A, B, C be arbitrary.  We have three easy containments.\n\n1.  A ⊆ A ∪ B ∪ C, so by (iii) A★ ⊆ (A ∪ B ∪ C)★.  \n With (ii) this yields (A★)★ = A★ ⊆ (A ∪ B ∪ C)★.\n\n2.  B ⊆ A ∪ B ∪ C, so again by (iii) B★ ⊆ (A ∪ B ∪ C)★.\n\n3.  Trivially C ⊆ A ∪ B ∪ C, hence by (i)  C ⊆ (A ∪ B ∪ C)★.\n\nPutting the three inclusions together gives\n\n (A★)★ ∪ B★ ∪ C ⊆ (A ∪ B ∪ C)★,\n\nwhich is exactly axiom (1).\n\nTherefore (1) and (i)–(iii) are equivalent, and the cloak operation enjoys inflation, idempotence and monotonicity as required. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.404804",
        "was_fixed": false,
        "difficulty_analysis": "• Several independent sets (A, B, C) now interact in a single axiom instead of just two; tracking how each of the three pieces behaves under the cloak map forces considerably more bookkeeping.\n\n• Both the cloak of a set (B★) and an uncloaked set (C) appear on the left of the axiom, so the proof must handle the interaction between “closed” and “raw” data simultaneously.\n\n• Deriving idempotence is subtler: it comes from recognising that the same axiom, specialised with B = C = A, simultaneously furnishes two opposite inclusions.  In the original kernel this step required only two sets; here one must keep three coordinated.\n\n• Monotonicity can no longer be read off directly from a single substitution.  One has to choose B and C cunningly (one cloaked, one uncloaked) so that A★ shows up explicitly inside the big union given by the axiom.\n\n• The converse direction likewise demands a triple-layer argument, feeding each of A★, B★ and C into properties (i)–(iii) separately before recombining them.\n\n• Overall the problem contains more variables, more cases to analyse, and a less obvious route to idempotence and monotonicity, making the reasoning chain distinctly longer and more intricate than in either the original problem or the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}