ada: Predefined arithmetic operators incorrectly treated as directly visible

In some cases, a predefined operator (e.g., the "+" operator for an
integer type) is incorrectly treated as being directly visible when
it is not. This can lead to both accepting operator uses that should
be rejected and also to incorrectly rejecting legal constructs as ambiguous
(for example, an expression "Foo + 1" where Foo is an overloaded function and
the "+" operator is directly visible for the result type of only one of
the possible callees).

gcc/ada/

	* sem_ch4.adb (Is_Effectively_Visible_Operator): A new function.
	(Check_Arithmetic_Pair): In paths where Add_One_Interp was
	previously called unconditionally, instead call only if
	Is_Effectively_Visible_Operator returns True.
	(Check_Boolean_Pair): Likewise.
	(Find_Unary_Types): Likewise.

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