For which one of the following chains does the conclusion R ≥ T FAIL to follow?

R ≥ T needs an unbroken run of same-sense links from R down to T. 'R ≥ S ≥ T' has exactly that and composes to R ≥ T. 'R > S = T' gives R > T, and R > T certainly implies R ≥ T, so it follows there too. 'R = S ≥ T' passes R's position across the equality and gives R ≥ T. Only 'R ≥ S ≤ T' breaks the pattern: both links point down towards S, so S is the smallest and R and T are never compared. R = 9, S = 1, T = 2 gives R ≥ T, but R = 3, S = 1, T = 40 satisfies the chain and gives R < T, so the conclusion is not forced. That reversal at S is the single feature that separates it from the other three.