Does `steady_clock::now()` have total order? [duplicate]

Within one thread, steady_clock::now() is guaranteed to return monotonically increasing values. How does this interact with memory ordering and reads observed by multiple threads?

atomic<int> arg{0};
steady_clock::time_point a, b, c, d;
int e;
thread t1([&](){
    a = steady_clock::now();
    arg.store(1, memory_order_release);
    b = steady_clock::now();
});
thread t2([&](){
    c = steady_clock::now();
    e = arg.load(memory_order_acquire);
    d = steady_clock::now();
});
t1.join();
t2.join()
assert(a <= b);
assert(c <= d);

Here's the important bit:

if (e) {
    assert(a <= d);
}

Can that assert ever fail? Or have I misunderstood something about acquire release memory order?