Follow

ロビンソンの$\mathbf{Q}$については、$\forall m\forall n[\mathbf{Q}\vdash\bar{m}\bar{n}=\bar{n}\bar{m}]$ だが $\mathbf{Q}\not\vdash\forall x\forall y\:xy=yx$ です。

· · Web · 0 · 1 · 2
Sign in to participate in the conversation
Mathtodon

A Mastodon instance named Mathtodon, where you can post toots with beautiful mathematical formulae in TeX/LaTeX style.