Download LSAT PrepTest 11 by LSAC PDF

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Proof If x expresses the same value as y, then both x and y will simplify to the same simple expression, call it e . s Let v = e . Thus v will also simplify to e , and so v can be reached from either x ory by taking steps to e and then retracing the simplification of v. s s s Thus x = v and PROCEDURAL THEOREMS 5 - 7 Therefore, by the convention of substitution, both x and y may be changed for an identical expression v in each case. But x and y can be any equivalent expressions. Therefore, expressions of the same value can be identified.

In case 0 n+ C3. In case 1 ~a~] \ r = ar CI CI. In case 2 71 T i l r = ~br~] ~a~r~\\ J2. In case more than 2 . . 71 71 T i l r .. 71 = ~ar\ . ~cr\ ~br\ ~a~r\\ J2 (as often as necessary) CI (as often as before). This completes the proof. Theorem 11 The scope of C8 can be extended as in T10. 7 ] 71 7 ] . . j 71 7] ~a~] ~Ì7\ ~c7]. ] = Theorem 12 The scope of C9 can be extended as in T10. TV7\\ TTTÌÌ = 7] ab . . TTTI tT71 .. xy Proofs of T i l and T12 follow from demonstrations as in C8 and C9, using T10 instead of J2.

N n Call a space reached by the greatest number of inwards crossings from 5 a deepest space in a. Call the space reached by no crossing from s the shallowest space in a. Thus so = s. Let any cross standing in any space in a cross c be said to be contained in c. Let any cross standing in the shallowest space in c be said to stand under, or to be covered by, c. Unwritten cross Suppose any so to be surrounded by an unwritten cross. Call the crosses standing under any cross c, written or unwritten, the crosses pervaded by the shallowest space in c.

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