Quick A: No, because acq_rel doesn't prevent reordering a store to x followed by a load from y... seq_cst does.

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Why isn't a C++11 acquire_release fence enough for Dekker synchronization?

The failure of Dekker-style synchronization is typically explained with reordering of instructions. I.e., if we write

atomic_int X;
atomic_int Y;
int r1, r2;
static void t1() {
    X.store(1, std::memory_order_relaxed)
    r1 = Y.load(std::memory_order_relaxed);
}
static void t2() {
    Y.store(1, std::memory_order_relaxed)
    r2 = X.load(std::memory_order_relaxed);
}

Then the loads can be reordered with the stores, leading to r1==r2==0.

I was expecting an acquire_release fence to prevent this kind of reordering:

static void t1() {
    X.store(1, std::memory_order_relaxed);
    atomic_thread_fence(std::memory_order_acq_rel);
    r1 = Y.load(std::memory_order_relaxed);
}
static void t2() {
    Y.store(1, std::memory_order_relaxed);
    atomic_thread_fence(std::memory_order_acq_rel);
    r2 = X.load(std::memory_order_relaxed);
}

The load cannot be moved above the fence and the store cannot be moved below the fence, and so the bad result should be prevented.

However, experiments show r1==r2==0 can still occur. Is there a reordering-based explanation for this? Where's the flaw in my reasoning?