Either something is A or B or everything is bC or d.
This conclusion may also be expressed in the form
If everything is ab, then every c is d.
[518] We cannot, if we are to be left with an equivalent proposition, express the first three of these alternants in a non-compound form. See sections [477], [479].
510. Six children, A, B, C, D, E, F are required to obey the following rules: (1) on Monday and Tuesday no four can go out together; (2) on Thursday, Friday, and Saturday no three can stay in together; (3) on Tuesday, Wednesday, and Saturday, if B and C are together, then A, B, E, and F must be together; (4) on Monday and Saturday B cannot go out unless either D, or A, C, and E stay at home. A and B are first to decide what they will do, and C makes his decision before D, E, and F. Find (α) when C must go out, (β) when he must stay in, and (γ) when he may do as he pleases. [Johns Hopkins Studies, p. 58.]
Let A = case in which A goes out, a = that in which he stays in, &c.
Then the premisses are as follows:
(1) On Monday and Tuesday,—three at least must stay in ;
(2) On Thursday, Friday, and Saturday,—no three can stay in together ;
(3) On Tuesday, Wednesday, and Saturday,—Every case is ABEF or abef or Bc or bC ;
(4) On Monday and Saturday,—Every case is ace or b or d.
In order to solve the problem, we must combine the possibilities for each day, then eliminate D, E, and F, and find in what ways the movements of A and B determine those of C.
(i) On Monday,—we have Every case is ace or b or d, combined with the condition that three at least must stay in. One alternant therefore is def without further condition, and it follows that we can determine no independent relation between A, B, and C.
Hence on Monday C may do as he pleases.
(ii) On Tuesday,—we have Every case is ABEF or abef or Bc or bC, combined with the condition that three at least must stay in. Therefore, Every case is abef or Bc or bC ;[519] and eliminating D, E, and F, Every case is ab or Bc or bC.
[519] The two alternants Bc and bC might here be made more determinate, thus, aBcd or aBce or aBcf or Bcde or Bcdf or Bcef and abCd or abCe or abCf or bCde or bCdf or bCef. But since we know that we are going on immediately to eliminate d, e, and f, it is obvious, even without writing them out in full, that these more determinate expressions will at once be reduced again to Bc and bC simply.
521 Hence it follows that on Tuesday (α) if A goes out while B stays in, C must go out, and (β) if B goes out, C must stay in.
(iii) On Wednesday,—Every case is ABEF or abef or Bc or bC ; or, eliminating D, E, and F, Every case is AB or ab or Bc or bC. Therefore, All Ab is C and All aB is c.
Hence on Wednesday (α) if A goes out while B stays in, C must go out, and (β) if A stays in while B goes out, C must stay in.
(iv) On Thursday and Friday,—the only condition is that no three can stay in together.
Hence on Thursday and Friday if A and B both stay in, C must go out.
(v) On Saturday,—Every case is ABEF or abef or Bc or bC ; also Every case is ace or b or d. Combining these premisses, Every case is ABdEF or abef or aBce or Bcd or bC. But we have the further condition that no three can stay in together. Therefore, Every case is ABdEF or ABcdEF or AbCDE or AbCDF or AbCEF or bCDEF. Therefore, eliminating D, E, and F, Every case is AB or bC.
Hence on Saturday if B stays in, C must go out.
511. Given (1) All P is QR, (2) All p is qr ; shew that (3) All Q is PR, (4) All R is PQ. [K.]