# Conjunction and disjunction

LEMMA    $$P \wedge (Q \vee R) \vdash P \wedge Q \vee P \wedge R$$

Proof available

LEMMA    $$P \wedge Q \vee P \wedge R \vdash P \wedge (Q \vee R)$$

Proof available

THEOREM

Conjunction is left-distributive over disjunction.

$$P \wedge (Q \vee R) \dashv\vdash P \wedge Q \vee P \wedge R$$

Proof available

LEMMA    $$(Q \vee R) \wedge P \vdash Q \wedge P \vee R \wedge P$$

Proof available

LEMMA    $$Q \wedge P \vee R \wedge P \vdash (Q \vee R) \wedge P$$

Proof available

THEOREM

Conjunction is right-distributive over disjunction.

$$(Q \vee R) \wedge P \dashv\vdash Q \wedge P \vee R \wedge P$$

Proof available

LEMMA    $$P \vee Q \wedge R \vdash (P \vee Q) \wedge (P \vee R)$$

Proof available

LEMMA    $$(P \vee Q) \wedge (P \vee R) \vdash P \vee Q \wedge R$$

Proof available

THEOREM

Disjunction is left-distributive over conjunction.

$$P \vee Q \wedge R \dashv\vdash (P \vee Q) \wedge (P \vee R)$$

Proof available

LEMMA    $$Q \wedge R \vee P \vdash (Q \vee P) \wedge (R \vee P)$$

Proof available

LEMMA    $$(Q \vee P) \wedge (R \vee P) \vdash Q \wedge R \vee P$$

Proof available

THEOREM

Disjunction is right-distributive over conjunction.

Proof available

THEOREM

Conjunction absorbs disjunction.

$$P \wedge (P \vee Q) \dashv\vdash P$$

Proof available

THEOREM

Disjunction absorbs conjunction.

$$P \vee (P \wedge Q) \dashv\vdash P$$

Proof available