Expose `resolvedModules` property of `SourceFile`

TypeScript Version: 3.6.4

Code

import * as ts from 'typescript'

function createCompilerHost(): ts.CompilerHost {
  function fileExists(fileName: string): boolean {
    return ts.sys.fileExists(fileName)
  }

  function readFile(fileName: string): string | undefined {
    return ts.sys.readFile(fileName)
  }

  function getSourceFile(
    fileName: string,
    languageVersion: ts.ScriptTarget,
  ) {
    const sourceText = ts.sys.readFile(fileName)
    return sourceText !== undefined
      ? ts.createSourceFile(fileName, sourceText, languageVersion)
      : undefined
  }

  function resolveModuleNames(moduleNames) {
    return moduleNames.map(() => undefined)
  }

  return {
    getSourceFile,
    getDefaultLibFileName: () => 'lib.d.ts',
    writeFile: (fileName, content) => ts.sys.writeFile(fileName, content),
    getCurrentDirectory: () => ts.sys.getCurrentDirectory(),
    getDirectories: path => ts.sys.getDirectories(path),
    getCanonicalFileName: fileName =>
      ts.sys.useCaseSensitiveFileNames ? fileName : fileName.toLowerCase(),
    getNewLine: () => ts.sys.newLine,
    useCaseSensitiveFileNames: () => ts.sys.useCaseSensitiveFileNames,
    fileExists,
    readFile,
    resolveModuleNames,
  }

}

const options = {
  module: ts.ModuleKind.ES2015,
  target: ts.ScriptTarget.ES5,
}
const host = createCompilerHost()
const sourceFiles = ts.createProgram(['./test.ts'], options, host).getSourceFiles()
const matchsFile = sourceFiles
  .filter(s => {
    return s.resolvedModules.has('typescript')
  })

Expected behavior:
Property 'resolvedModules' should exist on type 'SourceFile'

Actual behavior:

Start by locating the public SourceFile type and the createProgram().getSourceFiles() path shown in the reproduction. Check how resolvedModules is represented internally, then verify that the property is exposed on SourceFile and that the provided TypeScript example can access it.