Question

Every phone has a charming port.
Some phones have a headphone port.
Therefore some phones have both a charging and a headphone port.

predicate and proof?

Answer 1

The predicate and the proof is:

Let us consider P: Phone Q: charging port R : headphone port

The given predigates can be written as fron (PC2)— Qc+)2 (34) (PC7) + RC2)] : (31) (QCz)+ Rcm)) we have to prove fux) Cpca) + Q(x) ^ (ar) (P() R(x)] → (900 (Q(x)+R (20)

proof: 1. Un) P(1) - Q(x) 2. (Jx) P(X) + RCA) 3. P(S) 4. PCS) + Q(8) 5. Q (5) 6. PCS) – R(S) 7. R(S) 8. QC) +R(S) B 9. (IV) (ca) +RC=)) (hypothesis) (hypothesis) (using ) cusing 1) (using 3,4) cusing 2) cusing 3,6) (using 5,7) (using 8) Hence $xx) (CPC) Qc)) ^ (30) CPC2) +R(W) - (1x) (QCx) +R (m))