I am practicing my structural induction proofs. Suppose you have the following logic:

$\phi ::== True \ \vert \ \langle \rangle \phi \ \vert \ \phi_1 \land \phi_2$

Where $s \models \langle a \rangle \phi$ means there is a transition $ s \xrightarrow{a} s'$ such that $s'\models \phi$

Consider the following transition systems $S$ and $T$: $$ \require{AMScd} S: \hspace{10mm} \begin{CD} @. s_0 \\ @. @V{a}VV \\ s_2 @<{b}<< s_1 @>{c}>> s_3 \end{CD} \hspace{10mm} T: \hspace{10mm} \begin{CD} t_1 @<{a}<< t_0 @>{a}>> t_2 \\ @V{b}VV @. @V{c}VV\\ t_3 @. @. t_4\\ \end{CD} \\[30pt] $$ Claim: $T \models \phi \implies S \models \phi$

Proof: by structural induction on $\phi$

Base Case: $T \models True \implies S \models True $

Induction Step: Assume $T \models \phi \implies S \models \phi$

Need to show that $T \models \phi_1 \land \phi2 \implies S \models \phi_1 \land \phi2$

$T \models \phi_1 \land \phi_2 \iff T \models \phi_1$ and $T \models \phi_2 \implies S \models \phi_1$ and $S \models \phi_2 \iff S \models \phi_1 \land \phi_2$

Need to show $T \models \langle \rangle \phi \implies S \models \langle \rangle \phi$

From the transition systems you get:

$t_1 \models \langle b \rangle True \hspace{10pt} t_2 \models \langle c \rangle True \hspace{10pt} t_0 \models \langle a \rangle \langle b \rangle True \hspace{10pt} t_0 \models \langle a \rangle \langle c \rangle True$

$ t_0 \models \langle a \rangle \langle b \rangle True \land \langle a \rangle \langle c \rangle True $

$s_1 \models \langle b \rangle True \hspace{10pt} s_1 \models \langle c \rangle True \hspace{10pt} s_1 \models \langle b \rangle True \land \langle c \rangle True \hspace{10pt} s_0 \models \langle a \rangle \langle b \rangle True \hspace{10pt} s_0 \models \langle a \rangle \langle c \rangle True$

$s_0 \models \langle a \rangle (\langle b \rangle True \land \langle c \rangle True) \hspace{10pt} s_0 \models \langle a \rangle \langle b \rangle True \land \langle a \rangle \langle c \rangle True$

Hence $T \models \phi \implies S \models \phi \hspace{10pt} \blacksquare$

Does this look correct?


As a preliminary observation, you should mention that $T \models \phi$ is defined by $t_0 \models \phi$. In general, with multiple initial states, one universally quantifies over them. This is customary, but it makes your post more self-contained.

Your proof enumerates all formulae that are true at each state of each structure. I'm not sure I'd call it a proof by induction. Every state models $True$, which you don't mention explicitly, but clearly assume. (Note that your base case is something slightly different.) But apart from that, your enumeration is complete and correct.

Of course, the only bisimilar pairs of states are pairs of terminal states like $(t_3,s_3)$. So, the conclusion is that the structures are not bisimilar and not even simulation equivalent, but you probably knew that what you are establishing is that $S$ simulates $T$.


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