Intutionistic Logic: a logic that does not assume the law of excluded middle, i.e. P \lor \neg P is not always true. It is also called constructive logic because it requires a constructive proof of P or \neg P to conclude P \lor \neg P.
\begin{align*}
&\vdash P \to A \;\mathbf{says}\; P \\
&\vdash A \;\mathbf{says}\; (P \to Q) \to (A \;\mathbf{says}\; P \to A \;\mathbf{says}\; Q) \\
&\vdash A \;\mathbf{says}\; (A \;\mathbf{says}\; P) \to A \;\mathbf{says}\; P \\
&\not\vdash (A \;\mathbf{says}\; p) \to p
\end{align*}
\begin{align*}
&\frac{}{\Gamma, P \vdash P} \text{id} \\
\frac{\Gamma, P \vdash Q}{\Gamma \vdash P \to Q} {\to R}
\qquad&
\frac{\Gamma \vdash P \quad \Gamma, Q \vdash \delta}{\Gamma, P \to Q \vdash \delta} {\to L} \\
\frac{\Gamma \vdash P \quad \Gamma \vdash Q}{\Gamma \vdash P \land Q} {\land R}
\qquad&
\frac{\Gamma, P, Q \vdash \delta}{\Gamma, P \land Q \vdash \delta} {\land L} \\
\frac{\Gamma \vdash P(y) \quad y \notin (\Gamma, \forall x.\, P(x))}{\Gamma \vdash \forall x.\, P(x)} {\forall R^y}
\qquad&
\frac{\Gamma, P(c) \vdash \delta}{\Gamma, \forall x.\, P(x) \vdash \delta} {\forall L} \\
\frac{\Gamma \vdash P}{\Gamma \vdash P \lor Q} {\lor R_1}
\quad
\frac{\Gamma \vdash Q}{\Gamma \vdash P \lor Q} {\lor R_2}
\qquad&
\frac{\Gamma, P \vdash \delta \quad \Gamma, Q \vdash \delta}{\Gamma, P \lor Q \vdash \delta} {\lor L}
\end{align*}
\begin{align*}
\frac{}{\Gamma, P \vdash P} \text{id} \\
\frac{\Gamma, P \vdash Q}{\Gamma \vdash P \to Q} {\to R}
\qquad&
\frac{\Gamma \vdash P \quad \Gamma, Q \vdash \delta}{\Gamma, P \to Q \vdash \delta} {\to L} \\
\frac{\Gamma \vdash P \quad \Gamma \vdash Q}{\Gamma \vdash P \land Q} {\land R}
\qquad&
\frac{\Gamma, P, Q \vdash \delta}{\Gamma, P \land Q \vdash \delta} {\land L} \\
\frac{\Gamma \vdash P(y) \quad y \notin (\Gamma, \forall x.\, P(x))}{\Gamma \vdash \forall x.\, P(x)} {\forall R^y}
\qquad&
\frac{\Gamma, P(c) \vdash \delta}{\Gamma, \forall x.\, P(x) \vdash \delta} {\forall L} \\
\frac{\Gamma \vdash P}{\Gamma \vdash P \lor Q} {\lor R_1}
\quad
\frac{\Gamma \vdash Q}{\Gamma \vdash P \lor Q} {\lor R_2}
\qquad&
\frac{\Gamma, P \vdash \delta \quad \Gamma, Q \vdash \delta}{\Gamma, P \lor Q \vdash \delta} {\lor L} \\
\frac{\Gamma \vdash A \;\mathbf{aff}\; P}{\Gamma \vdash (A \;\mathbf{says}\; P)} {\mathbf{says}R}
\qquad&
\frac{\Gamma, P \vdash A \;\mathbf{aff}\; Q}{\Gamma, A \;\mathbf{says}\; P \vdash A \;\mathbf{aff}\; Q} {\mathbf{says}L} \\
&\frac{\Gamma \vdash P}{\Gamma \vdash A \;\mathbf{aff}\; P} \;\mathbf{aff}
\end{align*}
Notice R_1 and R_2 makes proof search more difficult. Above rules are not invertible.