Let's assume I have Invert//Invent targeting Erratic Cyclops by the first half. What happens - will he become 8/7 or will he become 15/0 and then die painfully?
Power/Toughness switching effects are always applied last. That means you first determine what is the creature's P/T values using all effects except the switch, and only then you switch them.
Now, as Invert//Invent does not have Fuse, you can cast only one side. While on the stack as a spell, it has only the characteristics and values from the one side that was cast.
So if you cast it as Invert targeting Cyclops, it is just a CMC 1 spell. Cyclops gets only +1/+0 from its own ability.
So, the correct answer is that Cyclops becomes 8/1 until end of turn.
Note that in addition to what willdice said this will continue to happen until end of turn. If you cast Divination after you will have an 8/4 Erratic Cyclops as each additional spell cast will be calculated before the switch happens at the end.
(to expand on Genini2's comment)
Whenever anything happens that changes an object's characteristics, that object's final appearance must be re-evaluated from scratch. This means you're expected to sift through all layers of rule 613.
Of course, we seldom have to go through the whole process, as many layers remain untouched.
But this means, as Willdice mentioned, that 'Power/Toughness switching effects are always applied last', since this is the very last layer of the list.
Therefore, increase/decrease to PT is always calculated before the switch, even if the switch effect was played earlier.
613.1. (…) all applicable continuous effects are applied in a series of layers in the following order (…)
In this case where an Erratic Cyclops becomes an 8/1 after using invert, what happens at the end of the turn when Erratic Cyclops's power is supposed to reset to zero?
In this case where an Erratic Cyclops becomes an 8/1 after using invert, what happens at the end of the turn when Erratic Cyclops's power is supposed to reset to zero?
Since both effects, power pump and p/t switch, have the same duration, they stop at the same time. The Cyclops is just immediately a 0/8, never a 1/8 or 8/0.
In this case where an Erratic Cyclops becomes an 8/1 after using invert, what happens at the end of the turn when Erratic Cyclops's power is supposed to reset to zero?
The effects of Invert and Erratic Cyclops's last ability both last "until end of turn" (C.R. 611.2a, 108.1), so if both effects are active, both will end during the cleanup step, and they will end simultaneously (C.R. 514.2). Thus, in this scenario, Erratic Cyclops will go back to being 0/8 — there will be no time when Erratic Cyclops's toughness is 0 here.
Also, even if those effects ended sequentially, it is not having 0 toughness that kills the creature, it is having 0 toughness when state based actions are checked that kills it. A creature can have 0 or less toughness for a while, or a player can have 0 or less life for a while, etc., without issue. So long as that while is in between two checks of state based actions, all is fine. State based actions are the cleanup crew of the game, they repeatedly check that everything is ok during the turn, and correct what is wrong. But they don't do so continuously.
Now, as Invert//Invent does not have Fuse, you can cast only one side. While on the stack as a spell, it has only the characteristics and values from the one side that was cast.
So if you cast it as Invert targeting Cyclops, it is just a CMC 1 spell. Cyclops gets only +1/+0 from its own ability.
So, the correct answer is that Cyclops becomes 8/1 until end of turn.
Whenever anything happens that changes an object's characteristics, that object's final appearance must be re-evaluated from scratch. This means you're expected to sift through all layers of rule 613.
Of course, we seldom have to go through the whole process, as many layers remain untouched.
But this means, as Willdice mentioned, that 'Power/Toughness switching effects are always applied last', since this is the very last layer of the list.
Therefore, increase/decrease to PT is always calculated before the switch, even if the switch effect was played earlier.
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