How to get this Qt state machine to work?

I have two widgets that can be checked, and a numeric entry field that should contain a value greater than zero. Whenever both widgets have been checked, and the numeric entry field contains a value greater than zero, a button should be enabled. I am struggling with defining a proper state machine for this situation. So far I have the following:

Code:

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QStateMachine *machine = new QStateMachine(this);
 
QState *buttonDisabled = new QState(QState::ParallelStates);
buttonDisabled->assignProperty(ui_->button, "enabled", false);
 
QState *a = new QState(buttonDisabled);
QState *aUnchecked = new QState(a);
QFinalState *aChecked = new QFinalState(a);
aUnchecked->addTransition(wa, SIGNAL(checked()), aChecked);
a->setInitialState(aUnchecked);
 
QState *b = new QState(buttonDisabled);
QState *bUnchecked = new QState(b);
QFinalState *bChecked = new QFinalState(b);
employeeUnchecked->addTransition(wb, SIGNAL(checked()), bChecked);
b->setInitialState(bUnchecked);
 
QState *weight = new QState(buttonDisabled);
QState *weightZero = new QState(weight);
QFinalState *weightGreaterThanZero = new QFinalState(weight);
weightZero->addTransition(this, SIGNAL(validWeight()), weightGreaterThanZero);
weight->setInitialState(weightZero);
 
QState *buttonEnabled = new QState();
buttonEnabled->assignProperty(ui_->registerButton, "enabled", true);
 
buttonDisabled->addTransition(buttonDisabled, SIGNAL(finished()), buttonEnabled);
buttonEnabled->addTransition(this, SIGNAL(invalidWeight()), weightZero);
 
machine->addState(registerButtonDisabled);
machine->addState(registerButtonEnabled);
machine->setInitialState(registerButtonDisabled);
machine->start();

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The problem here is that the following transition:

Code:

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buttonEnabled->addTransition(this, SIGNAL(invalidWeight()), weightZero);

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causes all the child states in the registerButtonDisabled state to be reverted to their initial state. This is unwanted behaviour, as I want the a and b states to remain in the same state.

How do I ensure that a and b remain in the same state? Is there another / better way this problem can be solved using state machines?

Note. There are a countless (arguably better) ways to solve this problem. However, I am only interested in a solution that uses a state machine. I think such a simple use case should be solvable using a simple state machine, right?