Concern the following axiom set in which tree and row are un

Concern the following axiom set, in which \'\'tree\" and \'\'row\" are undefined terms. There exists at least 1 tree. Every row has exactly 2 trees. Every tree belongs to at least 1 row. Any 2 trees have exactly 1 row in common. For every row R there is exactly 1 other row having no trees in common with R. Prove Theorems 1 through 8. Theorem: There exists at least 1 row. Theorem: There exist at least 2 rows. Theorem: There exist at least 4 trees. Theorem: Every tree belongs to at least 2 rows. Theorem: There exist at least G trees. Theorem: There exist exactly 4 trees. Theorem: There exist exactly 6 rows. Theorem: Each tree belongs to exactly 4 rows.

Solution

1) From Axiom A we have, There exists at least 1 tree and from Axiom C, this tree belongs to at least 1 row means that the existing tree should belong to a row. This implies there exist at least one row (to contain the existing tree)

2) From theorem 1, jthere exists at least 1 row say R1 and form Axiom E, there is exactly 1 other row say R2 having no trees in common with R1. That is there exists at least 2 rows.

3) From theorem 2, we have at least 2 rows and of them at least two are disjoint (that is no tree is common to them) From Axiom B, every row has exactly 2 trees. So, the existing (at least) 2 rows contains (at least) 4 trees.

4) Consider a tree say T1 (i.e., existing by axiom 1). By Axiom C, this tree should belong to atleast one row say R1. From axiom E there exists another row say R2 having no trees in common with R1. Now consider a tree T2 from R2. From axiom D, T1 and T2 are common in exactly 1 row say R3. Therefore the tree T1 belongs to at least two rows R1 and R3. This is true for any tree. So every tree belongs to at least two rows.