Puzzle: Find a Still Life in B2e3/S1c23a

[Hunting]

Find a Still Life in B2e3/S1c23a without block or carrier!
This is just for fun. I think it should go to the sandbox.

[77topaz]

I am fairly certain that's actually impossible.

[Hunting]

I am fairly certain that's actually impossible.

So prove it! Just like proving there aren't any B3/S3 still life without block!

[Hunting]

(Updating)
I think I can prove that this puzzle cannot be solved
Assume (0,0) is on

Code: Select all

x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$A19D!

Think about 1c. if it is saved by the 1c rule, then:

Code: Select all

x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$DA$A19D!

That will create a born on (1,0) and (0,1). If (2,0) , (2,1) , (0,2) and (1,2) is on:

Code: Select all

x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$2A$D2A$ADA17D!

(2,1), (1,2) and (1,1) will die.
If (1,0) and (0,1) is on:

Code: Select all

x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$2A$2A18D!

、
A block appears.
So (1,1) and (0,0) can't be on at the same time.
If it is saved by the S2 rule, there are two possibilities, but (1,1) must be off, so (0,1) and (1,0) must be on.

Code: Select all

x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$AB$2A18D!

(0,1) and (1,0) already had two neighbour, so the blue cells must be off:

Code: Select all

x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$2B$A2B$2AB17D!

This will cause a B3a on (1,1), so (2,2) must be on
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