Gödel's incompleteness wrt weakend versions of ZFC

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Removing the axiom of infinity from ZFC leads to a theory that does not satisfy the hypotheses of Gödel's incompleteness theorems. This modified version of ZFC, referred to as ZFC-, cannot express all arithmetical truths and is effectively equivalent to Peano Arithmetic. Consequently, any subtheory of an incomplete theory remains incomplete. The discussion highlights that ZFC- cannot be both consistent and complete due to its foundational limitations. Overall, the implications of altering axioms in ZFC raise significant questions about the nature of mathematical completeness and consistency.
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Suppose for the sake of argument that we look at ZFC with the axiom of infinity removed.

http://en.wikipedia.org/wiki/Gödel's_incompleteness_theorems#First_incompleteness_theorem

http://en.wikipedia.org/wiki/Gödel's_incompleteness_theorems#Second_incompleteness_theorem

We would then be in a position where the hypotheses of Gödel's theorems are not satisfied, correct? Basically, I want to remove, for the sake of argument, a minimal amount of axioms of ZFC so that ZFC minus some axiom(s) leads to a theory that does not include arithmetical truths and is not capable of expressing elementary arithmetic.

Is it possible that ZFC- (my shorthand for ZFC with some axiom(s) removed) is consistent and complete?
 
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No, obviously any subtheory of an incomplete theory is also incomplete.

Also, ZFC without the axiom of infinity is basically Peano Arithmetic (or rather, replacing the Axiom of infinity by its negation yields a theory which is bi-interpretable with PA).
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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