Replace this sentence with:

When a sub-flow is explicitly configured with a channel() as the first element, the framework implicitly places a bridge() between the flow's input channel and that channel.

I am having trouble parsing this sentence. I can reword it if I understood what you are trying to say. Perhaps if you can provide an example flow, I can try to reword it.

What I want to say here that sub-flow's firs consumer is a subscriber to the channel to which original component is producing.
Imaging we have a filter with its discardChannel. There definitely should be something subscribing to that channel and, well, frankly we always call that as a sub-flow.
In case of DSL we just don't mention that channel in between explicitly to avoid boilerplate code, but it is there anyway by the framework. Just because the mentioned filter really expects from us a MessageChannel for its discardChannel option. So, or we use an inputChannel of the IntegrationFlow bean you provide for the sub-flow or we create a new one and start sub-flow from it letting that filter to produce to this channel when it discards messages.

Hope this is getting cleaner...

It's not clear what a subflow has to do with a filter's discard channel; perhaps a DSL snippet will help me.

Well, what I mean is like this:

.filter(p -> p instanceof String, e -> e
	.discardFlow(df -> ...))

where that sub-flow is connected to an internally created discardChannel for the MessageFilter.
That's the point where that bridge() comes to the scene when you try to use an explicit channel() in the beginning of that flow.
Just because there is already a channel - created by the framework and we can't determine this explicit channel() because this sub-flow has not been parsed yet, unlike externally configured @Bean IntegrationFlow.

Got it. My confusion was because you said discardChannel, not discardFlow.

How about...

In cases where the DSL supports a subflow configuration, where a channel is normally needed for the component being configured, and that subflow starts with a channel() element, the framework implicitly places a bridge() between the flow's input channel and that channel.

For example, in...

.filter(p -> p instanceof String, e -> e
	.discardFlow(df -> df
                         .channel(MessageChannels.queue())
                         ...)
...

... the framework can't wire a subflow into a filter, so a discardChannel bean is created and bridged to the queue channel at the start of the subflow.

no such bridge