Add language version check for compound assignment null propagation

Compound assignments with null conditional (?.) are only supported in C# 14+.
Added language version check to ensure the diagnostic is only offered when supported.
Added test to verify diagnostic is not offered for C# 13.

Co-authored-by: CyrusNajmabadi <4564579+CyrusNajmabadi@users.noreply.github.com>

Author: Copilot

src/Analyzers/CSharp/Tests/UseNullPropagation/UseNullPropagationTests.cs

@@ -3006,4 +3006,22 @@ static void M(C c)
                 }
             }
             """, languageVersion: LanguageVersion.CSharp14);
+
+    [Fact]
+    public Task TestIfStatement_CompoundAssignment_NotAvailableInCSharp13()
+        => TestMissingInRegularAndScriptAsync(
+            """
+            using System;
+
+            class C
+            {
+                event Action SomeEvent;
+
+                static void M(C c)
+                {
+                    if (c is not null)
+                        c.SomeEvent += () => { };
+                }
+            }
+            """, LanguageVersion.CSharp13);
 }

src/Analyzers/Core/Analyzers/UseNullPropagation/AbstractUseNullPropagationDiagnosticAnalyzer.cs

@@ -448,8 +448,9 @@ private static TExpressionSyntax RemoveObjectCastIfAny(
 
         // Check if node itself is a compound assignment. We do this by checking if the first child node
         // is on the left side of a compound assignment (which means the node is the compound assignment itself).
+        // Compound assignments with null conditional (?.) are only supported in C# 14+.
         var firstChild = node.ChildNodesAndTokens().FirstOrDefault().AsNode();
-        if (firstChild != null && syntaxFacts.IsLeftSideOfCompoundAssignment(firstChild))
+        if (firstChild != null && syntaxFacts.IsLeftSideOfCompoundAssignment(firstChild) && syntaxFacts.SupportsNullConditionalAssignment(node.SyntaxTree.Options))
         {
             syntaxFacts.GetPartsOfAssignmentExpressionOrStatement(node, out var left, out _, out _);
             return (TExpressionSyntax)left;