Prove that [4x)A(x)]' → (Vx)[A(x)]' is valid. We must prove the implication in each direction. a. [ax)4(x)]' <→ (Vx)[A(x)]' The hypothesis alone gives us little to work with, so we introduce a (somewhat surprising) temporary hypothesis. A proof sequence is 1. [Ex)A(x)]' 2. A(x) 3. (Ex)A(x) hyp temporary hyp 2, eg 4. A(x) → (Ax)A(x) temporary hyp discharged 5. [A(x)]' 1, 4, mt 6. (Vx)[A(x)]' 5, ug b. (Vx)[4(x)]' → [x)A(x)]' This proof also requires a temporary hypothesis. It is even more surprising than case (a) because we assume the exact opposite of the conclusion we are trying to reach. 1. (Vx)[A(x)]' 2. (Ex)4(x) hyp temporary hyp 3. A(a) 2, ei 4. [A(a)]' 1, ui 5. [(Vx)[A(x)]'I' 6. (Ax)A(x) → [(Vx)[A(x)]]' temporary hyp discharged 3, 4, inc 7. [((Vx)[A(x)])]' 1, dn 8. [Ex)4(x)]' 6, 7, mt

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A: The solution is given below with proper explanation

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Longest Common Subsequence

The Longest Common Subsequence (LCS) is defined as the longest subsequence that is shared by all of the given sequences, assuming that the subsequence's elements do not have to be in consecutive locations in the original sequences. It is used in the areas like bioinformatic, revision control systems like SVN and Git.