Take 2, Part 2
To show that the higher-degree containment logic described earlier has the desired relevant properties:
To show \(A\rightarrow B\vDash(C\rightarrow A)\rightarrow(C\rightarrow B)\):
Suppose \(\mathfrak{M}\vDash A\rightarrow B\) but \(\mathfrak{M}\nvDash(C\rightarrow A)\rightarrow(C\rightarrow B)\). Then there exists a \(\gamma\) such that \(X\in \bar{V}^{+}_{\gamma}(C\rightarrow A)\) but \(X\notin \bar{V}^{+}_{\gamma}(C\rightarrow B)\). As \(X\in \bar{V}^{+}_{0}(A\rightarrow B)\), any \(\pi\in\bar{V}^{+}_{\gamma}(A)\) is also a member of \(\bar{V}^{+}_{\gamma}(B)\), i.e., \(\bar{V}^{+}_{\gamma}(A)\subseteq\bar{V}^{+}_{\gamma}(B)\). The semantic constraint implies that \(\bar{V}^{+}_{\gamma}(C\rightarrow A)\subseteq\bar{V}^{+}_{\gamma}(C\rightarrow B)\). [Note that trivially \(\bar{V}^{+}_{\gamma}\upharpoonright X=\bar{V}^{+}_{\gamma}\).] That \(X\in \bar{V}^{+}_{\gamma}(C\rightarrow A)\) thus entails that \(X\in \bar{V}^{+}_{\gamma}(C\rightarrow B)\). Contradiction.
To show \(A\leftrightarrow B\vDash(B\rightarrow C)\rightarrow(A\rightarrow C)\):
Suppose \(\mathfrak{M}\vDash A\leftrightarrow B\) but \(\mathfrak{M}\nvDash(B\rightarrow C)\rightarrow(A\rightarrow C)\). Then for some \(\gamma\), \(X\in \bar{V}^{+}_{\gamma}(B\rightarrow C)\) but \(X\notin \bar{V}^{+}_{\gamma}(A\rightarrow C)\). Since \(X\in \bar{V}^{+}_{0}(A\leftrightarrow B)\), \(\bar{V}^{+}_{\gamma}(A)=\bar{V}^{+}_{\gamma}(B)\), whence by the constraint \(\bar{V}^{+}_{\gamma}(B\rightarrow C)=\bar{V}^{+}_{\gamma}(A\rightarrow C)\). This entails that \(X\in \bar{V}^{+}_{\gamma}(A\rightarrow C)\). Contradiction.
To show \(A\rightarrow B\nvDash(B\rightarrow C)\rightarrow(A\rightarrow C)\):
We show this by countermodel. Let \(\mathfrak{M}=(X,\Pi,V)\) where \(X=\lbrace a\rbrace\), \(\Pi=\lbrace X\rbrace\), and functions \(V_{0}\) and \(V_{1}\) are such that the former is empty and the latter maps \(B\) and \(B\rightarrow C\) to \(\lbrace X\rbrace\) and all other formulae to \(\varnothing\). Then \(X\in\bar{V}^{+}_{0}(A\rightarrow B)\) vacuously as there are no states at which \(A\) holds. Also, \(X\in\bar{V}_{1}^{+}(B\rightarrow C)\) and \(X\in\bar{V}_{1}^{+}(A\rightarrow C)\). That \(\bar{V}_{1}^{+}(A)\neq\bar{V}_{1}^{+}(B)\) permits this evaluation. Hence, \(X\notin\bar{V}_{0}^{+}((B\rightarrow C)\rightarrow(A\rightarrow C))\), providing the necessary countermodel.
There is still much work to do. But it's a reasonable--and natural--first pass. (Although, strictly speaking, it's the second pass.)











