連結の理論
Theory of Concatination．$ \sf TC
参考文献
モチベーション
Def
連結の理論の言語$ \mathscr{L}_\mathsf{TC} := \lang \alpha,\beta; \frown; = \rang
$ \sf TCは以下の公理を持つ
1. $ \forall_{x,y,z}.\lbrack x \frown (y \frown z) = (x \frown y) \frown z \rbrack
2. $ \forall_{x,y,u,v}.\lbrack x \frown y = u \frown v \to \\ ((x = y \land y = v) \lor \exists_w.\lbrack (u = x \frown w \land w \frown v = y) \lor (x = u \frown w \land w \lor y = v) \rbrack ) \rbrack
example
$ \alpha\beta \frown \beta\alpha = \alpha\beta \frown \beta\alpha \implies \alpha\beta = \alpha\beta ~\&~ \beta\alpha = \beta\alpha
$ \alpha \frown \alpha\beta\alpha = \alpha\alpha\beta \frown \alpha \implies \alpha\alpha\beta = \alpha \frown \alpha\beta ~\&~ \alpha\beta \frown \alpha = \alpha\beta\alpha
このとき$ w = \alpha\beta
$ \alpha\beta\alpha \frown \beta = \alpha \frown \beta\alpha\beta \implies \alpha\beta\alpha = \alpha \frown \beta\alpha ~\&~ \beta\alpha \frown \beta = \beta\alpha\beta
このとき$ w = \beta\alpha
3. $ \forall_{x,y}.\lbrack \alpha \neq x \frown y \rbrack
4. $ \forall_{x,y}.\lbrack \beta \neq x \frown y \rbrack
5. $ \alpha \neq \beta
Thm
remark:
remark:
remark:
ところで$ \sf Q^-は$ \sf Qと互いに解釈可能
結果として$ \sf Qと$ \sf TCは互いに解釈可能．
その他の独立な証明は以下．