React Native，如何再次使redux状态为假

import { createStore } from 'redux';

const initialState = {
  isFalse: false,
};

function reducer(state = initialState, action) {
  switch (action.type) {
    case 'SET_FALSE':
      return { ...state, isFalse: true };
    default:
      return state;
  }
}

const store = createStore(reducer);

import React from 'react';
import { Provider } from 'react-redux';
import App from './App';
import store from './store';

export default function Main() {
  return (
    <Provider store={store}>
      <App />
    </Provider>
  );
}

import React from 'react';
import { View, Text } from 'react-native';
import { connect } from 'react-redux';

function MyComponent({ isFalse, setFalse }) {
  return (
    <View>
      <Text>isFalse: {isFalse.toString()}</Text>
      <Button title="Set False" onPress={setFalse} />
    </View>
  );
}

const mapStateToProps = state => ({
  isFalse: state.isFalse,
});

const mapDispatchToProps = dispatch => ({
  setFalse: () => dispatch({ type: 'SET_FALSE' }),
});

export default connect(mapStateToProps, mapDispatchToProps)(MyComponent);