# Is 1+1=2 logically equivalent to 99+1=100?

I am trying to understand logical equivalence.

From what I understand, two formulae are logically equivalent if they have the same truth values under all interpretations. So, $$x+1=y\dashv\vdash x+(2-1)=y$$ because under all interpretations of $$x$$ and $$y$$, the truth values are the same.

While the following biconditional isn't an equivalence: $$(x+1=2)\land y=y\iff(y+5=7)\land x=x$$ because there are values of $$x$$ and $$y$$ that would give the sentences different truth values.

But what if the formulae are atomic sentences which are both true or both false? Would these be equivalent sentences?

$$v(1+1=2)=\top$$

$$v(99+1=100)=\top$$

$$v(1+1=2\dashv\vdash 99+1=100)=?$$

Is 1+1=2 logically equivalent to 99+1=100?

• $2=1+1$ is more of a definition of the symbol "$2$" than being any declaration of substance. Commented Mar 6, 2023 at 7:59
• You may find this question useful.
– Jam
Commented Mar 8, 2023 at 7:48

The sentence $$1+1=2$$ is not logically equivalent with $$99 +1 =100$$.

To see this note the way logical equivalence is defined in model theory. In first-order logic two sentences $$\varphi, \psi$$ of some first-order language $$L$$ are logically equivalent iff $$M$$ models $$\varphi$$ exactly if $$M$$ models $$\psi$$, for every $$L$$-model $$M$$. The definition quantifies over every model for $$L$$, even if $$L$$ is the language of number theory. And there are no logical constraints on how to interpret non-logical constants like numerals or the addition symbol (besides assigning them objects of the right type).

To be more specific take $$L$$ to be the sublanguage of first-order Peano Arithmetic with the signature $$\{+,~ ', ~0 \}$$. For any non-zero natural number $$n$$ we may define the constant $$\underline{n} := 0' \ldots'$$ ($$n$$ bars). Let $$M = (\mathbb{N}, +_{M},~ '_{M},~ 0_{M})$$, where $$'_{M}$$ is the successor function, $$0_{M}$$ is the number $$0$$, but $$+_{M} = \{((1,1), 2)\}$$. Then it's immediate that $$M$$ models the $$L$$-sentence $$\underline{1}+\underline{1} = \underline{2}$$, but not $$\underline{99} +\underline{1} = \underline{100}$$.

Of course it is possible to restrict the class of $$L$$-models to those models which are isomorphic to the $$L$$-reduct of the standard model of arithmetic (setting aside non-standard models for the moment). Relative to this restricted class your example sentences indeed are always either both true or both false. But that does not amount to logical equivalence proper.

• Do you have anywhere where I can learn more about model theory? Commented Mar 6, 2023 at 3:00
• Good entry points are Chang / Keisler: 'Model Theory' as well as Wilfrid Hodges: 'A Shorter Model Theory'. These textbooks presuppose that you're familiar with cardinal arithmetic and abstract algebra. More gentle introductions are Manzano: 'Model Theory' and Kirby: 'An Invitation to Model Theory'. I Commented Mar 7, 2023 at 16:42

From what I understand, two formulae are logically equivalent if they have the same truth values under all interpretations. So, $$x+1=y\dashv\vdash x+(2-1)=y$$ because under all interpretations of $$x$$ and $$y$$, the truth values are the same.

No: if we reinterpret the operation - as returning the higher input value, then the left and right statements above are no longer equivalent, and thus not logically equivalent.

Is 1+1=2 logically equivalent to 99+1=100?

No: counterexample: redefine + to mean 'double the product of the inputs', then the left statement is true while the right statement is false.

• Wow, this is very clear. Thank you. Commented Mar 6, 2023 at 23:15
• @IsaacSechslingloff You're welcome! Since your query is essentially "Is a true statement equivalent to every true statement?", here's a companion answer that I wrote: Is a true statement negated by every false statement? Commented Mar 7, 2023 at 1:50