Enable button when toggle is check

Ionic Framework ionic-v3
1

Hello, I want enable register button only when the toglle condition of use is checked like this :

2018-01-13 14_20_13-WebJobbing - Ionic Lab

register.html :

<ion-item class="CGU" no-lines>
     <ion-label class="CGU">I accept the conditions of use</ion-label> 
     <ion-toggle color="secondary" [checked]="isToggled" (ionChange)="notify()"></ion-toggle>
</ion-item> 

<button ion-button class="login" color="bleu" type="submit" (click)="register()" [disabled]="isToggled != true">Register</button>

register.ts :

public isToggled: boolean = false;

constructor(...) {...}

notify() {
    this.isToggled = !this.isToggled;
  }

But it not working…

Nobody knows how to do it please?

2 
<ion-item class="CGU" no-lines>
     <ion-label class="CGU">I accept the conditions of use</ion-label> 
     <ion-toggle color="secondary" [(ngModel)]="agreements"></ion-toggle>
</ion-item> 

<button ion-button class="login" color="bleu" type="submit" (click)="register()" [disabled]="!agreements" Register</button>
3

Not working : Uncaught (in promise): Error: \n ngModel cannot be used to register form controls with a parent formGroup directive. Try using\n formGroup’s partner directive "formControlName" instead.

I think I can’t use Ngmodel because I already use that in the rest of my code for something else…

I think I must use [checked] and (ionChange)

4

ok it work like that :

<ion-item class="CGU" no-lines>
     <ion-label class="CGU">I accept the conditions of use</ion-label> 
     <ion-toggle color="secondary" [checked]="isToggled" (ionChange)="notify()"></ion-toggle>
</ion-item> 

<button ion-button class="login" color="bleu" type="submit" (click)="register()" [disabled]="!isToggled">Register</button>

quite simply…