The next theorem may be wrong but I submit it as an idea.
Theorem 8. Theory of series is wrong.
Let e(x) be a function symbol. Put k=Z/2Z={0",1"}.
Assume consistency of A. You may set up that ∀x(e(x)=1")∈A, 0"≠1"∈A.
Put Em=A∪{m∈d,d∈N, ∑i∈d e(i) ∈k} (m∈N).
Like the proof of theorem 1, ∪m∈N Em is consistent and has a countable
model M for which M |= m∈d for ∀m∈N.
you can set up that d={c1,c2,…} for M for M is a countable model.
e=∑i∈d e(i)=e(c1)+e(c2)+…(infinitely)…=1"+1"+…(infinitely)…∈k for M.
Seeing the sequence of symbols, you get M |= e=1"+e. M |= e+e=0" for M |= e∈k.
M |= 0"=e+e=1"+e+e=1"+0"=1" This is contradiction.♦
Theorem 8. Theory of series is wrong.
Let e(x) be a function symbol. Put k=Z/2Z={0",1"}.
Assume consistency of A. You may set up that ∀x(e(x)=1")∈A, 0"≠1"∈A.
Put Em=A∪{m∈d,d∈N, ∑i∈d e(i) ∈k} (m∈N).
Like the proof of theorem 1, ∪m∈N Em is consistent and has a countable
model M for which M |= m∈d for ∀m∈N.
you can set up that d={c1,c2,…} for M for M is a countable model.
e=∑i∈d e(i)=e(c1)+e(c2)+…(infinitely)…=1"+1"+…(infinitely)…∈k for M.
Seeing the sequence of symbols, you get M |= e=1"+e. M |= e+e=0" for M |= e∈k.
M |= 0"=e+e=1"+e+e=1"+0"=1" This is contradiction.♦