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I am currently reading "Boxes and Diamonds: An Open Introduction to Modal Logic". Here is the PDF to the book: https://builds.openlogicproject.org/courses/boxes-and-diamonds/bd-screen.pdf

One of the exercises reads the following: Prove Proposition 3.19 by proving, by induction on the complexity of C , that if K ⊢ A ↔ B then K ⊢ C [A/q ] ↔ C [B/q ].

Can someone please help me with proving this? I am just really curious how the proof actually would look.

Proposition 3.19.

Kylie
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1 Answers1

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Induction on $C$ means considering each possibility of $C$ as a formula, as given by Definition 1.2, and then applying the definition of substitution (1.4). For example: $\newcommand{\K}{{\bf K}}$

  1. If $C = \bot$ then $C[A/q] =\bot = C[B/q]$, so since $\K\vdash C[A/q]$ we know $\K\vdash C[B/q]$.
  2. If $C = p$ is a variable, then either
    1. $p = q$, in which case $C[A/q] = A$, and $C[B/q] = B$, but since $\K \vdash A\leftrightarrow B$ then $\K\vdash C[B/q]$
    2. $p \neq q$, in which case $C[A/q] = C = C[B/q]$, so $\K \vdash C[B/q]$.
  3. If $C = D \wedge E$, then $C[A/q] = D[A/q]\wedge E[A/q]$, but by the induction hypothesis, $\K\vdash D[A/q]$ implies $\K\vdash D[B/q]$, and similarly $\K\vdash E[A/q]$ implies $\K\vdash E[B/q]$, therefore $\K\vdash C[B/q]$. $$\vdots$$
Couchy
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