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ozarga
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Homework Statement
Hello, I'm having problems with the following exercise from my homework.
Proof that when A[itex]\subseteq[/itex]B, then it happens that C-B[itex]\subseteq[/itex]C-A
Homework Equations
The Attempt at a Solution
This is how I have been trying to solve it:
1. A[itex]\subseteq[/itex]B // Hyp
2. x[itex]\in[/itex]A[itex]\rightarrow[/itex]x[itex]\in[/itex]B //Element wise proof [1]
3. x[itex]\in[/itex]A [itex]\wedge[/itex] x[itex]\in[/itex]B // which rule allow me to do this? [2]
4. ( x[itex]\in[/itex]A [itex]\wedge[/itex] x[itex]\in[/itex]B) [itex]\vee[/itex] x[itex]\in[/itex]C // Addition [3]
5. (x[itex]\in[/itex]A[itex]\vee[/itex]x[itex]\in[/itex]C)[itex]\wedge[/itex](x[itex]\in[/itex]B[itex]\vee[/itex]x[itex]\in[/itex]C) // Distrivutive law [4]
6. x[itex]\in[/itex]A[itex]\vee[/itex]x[itex]\in[/itex]C // [itex]\wedge[/itex] OUT [5]
7. x[itex]\in[/itex]B[itex]\vee[/itex]x[itex]\in[/itex]C // [itex]\wedge[/itex] OUT [5]
8. y[itex]\in[/itex]C // Hyp (This step is the very confusing one. I'm assuming it exists an element of one set I have no information it actually exists)
9. y[itex]\notin[/itex]B // Modus Ponendo Tollens [7,8]
10. y[itex]\in[/itex]C[itex]\wedge[/itex]y[itex]\notin[/itex]B // [itex]\wedge[/itex] IN [8,9]
11. y[itex]\notin[/itex]A // Modus Ponendo Tollens [6,8]
12. y[itex]\in[/itex]C[itex]\wedge[/itex]y[itex]\notin[/itex]A // [itex]\wedge[/itex] IN [8,11]
13. y[itex]\in[/itex]C[itex]\wedge[/itex]y[itex]\notin[/itex]B [itex]\rightarrow[/itex] y[itex]\in[/itex]C[itex]\wedge[/itex]y[itex]\notin[/itex]A // CP [10, 12]
14. C-B [itex]\subseteq[/itex] C-A // Defs of Difference [13] & Element wise proof
Sorry for my poor english. Thanks in advance for your help.