In a code '✦' means 'is not greater than' and '△' means 'is greater than'. Three statements are given, the first in ordinary symbols and the other two in this code: P = Q, 'Q ✦ R', and 'R △ S'. Which of the following is definitely true?

'Is not greater than' is ≤, so the statements decode to P = Q, Q ≤ R and R > S. From the first two, the equality passes P's relation straight to Q, giving P ≤ R — that conclusion is safe. The tail is where the chain dies: Q ≤ R and R > S both point up towards R, so S sits below R but is never compared with P. P = Q = 2, R = 9, S = 1 gives P > S, while P = Q = 2, R = 9, S = 8 gives S > P; both satisfy every statement, so neither 'P > S' nor 'S > P' follows. 'R = P' would need the non-strict link Q ≤ R to be an equality, but P = Q = 2 with R = 9 satisfies the statements and refutes it.