Does indestructible work like this?
Asked by Scytec 9 years ago
So, the other day I was playing a game of EDH and my opponent played Nylea, God of the Hunt as a creature. I was holding Bile Blight . I had a Stromkirk Noble up to a 5/5 because of combat damage. So my question is this. He attacks with Nylea the next turn, I block with the noble assigning 5 damage to Nylea. I play Bile Blight giving her -3/-3 until end of turn. Nylea would still die correct? Because even though combat damage can not kill a creature with indestructible, it still takes the damage and then you can give them a-negative toughness effect to actually kill them correct?
GlistenerAgent says... Accepted answer #2
In your scenario, Nylea, God of the Hunt is still a 3/3 marked with lethal damage, which through her indestructibility ability would not cause her to die. For her to die, her toughness would have to actually become zero, possibly through a combination of Ulcerate and Bile Blight or something like that.
September 21, 2014 7:56 p.m.
Hmm...I thought the damage would still apply for the turn just in a lethal sense. Ok, thanks for the clarification.
September 21, 2014 9:08 p.m.
It is a common misconception that damage reduces toughness. It doesn't.
The game simply checks to see whether the amount of damage marked on a creature is equal to or greater than the toughness of the creature, and if it is, causes that creature to die due to lethal damage.
Nylea as a 6/6 does not become a 6/1 from the noble, thus becoming a 3/-2 and dying.
Instead, she is a 6/6 with 5 damage marked on her, which becomes a 3/3 with 5 damage marked on her. At this point the game would try to cause Nylea to die due to having lethal damage marked, but would fail because she is indestructible.
beckhr says... #1
No, to destroy an indestructible creature you need to take them to zero toughness through min using effects, such as Bile Blight or Infect. In the situation you describe, you're still killing her with damage.
September 21, 2014 7:52 p.m.