Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication
< Rule of Distribution | Disjunction Distributes over Conjunction | Left Distributive | Formulation 2
Jump to navigation
Jump to search
Theorem
- $\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }$
Proof 1
This proof is derived in the context of the following proof system: instance 1 of a Gentzen proof system.
By the tableau method:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\neg p, p, q$ | Axiom | ||||
2 | $\neg p, p \lor q$ | $\beta$-Rule: $\beta \lor$ | 1 | |||
3 | $\neg p, p, r$ | Axiom | ||||
4 | $\neg p, p \lor r$ | $\beta$-Rule: $\beta \lor$ | 3 | |||
5 | $\neg p, \paren {p \lor q} \land \paren {p \lor r}$ | $\alpha$-Rule: $\alpha \land$ | 2, 4 | |||
6 | $\neg q, \neg r, p, q$ | Axiom | ||||
7 | $\neg q, \neg r, p \lor q$ | $\beta$-Rule: $\beta \lor$ | 6 | |||
8 | $\neg q, \neg r, p, r$ | Axiom | ||||
9 | $\neg q, \neg r, p \lor r$ | $\beta$-Rule: $\beta \lor$ | 8 | |||
10 | $\neg q, \neg r, \paren {p \lor q} \land \paren {p \lor r}$ | $\alpha$-Rule: $\alpha \land$ | 7, 9 | |||
11 | $\neg \paren {q \land r}, \paren {p \lor q} \land \paren {p \lor r}$ | $\beta$-Rule: $\beta \land$ | 10 | |||
12 | $\neg \paren {p \lor \paren {q \land r} }, \paren {p \lor q} \land \paren {p \lor r}$ | $\alpha$-Rule: $\alpha \land$ | 5, 11 | |||
13 | $\paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }$ | $\beta$-Rule: $\beta \implies$ | 12 |
$\blacksquare$
Proof 2
This proof is derived in the context of the following proof system: Instance 2 of the Hilbert-style systems.
By the tableau method:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {q \land r} \implies q$ | Rule of Simplification | $q / p, r / q$ | |||
2 | $\paren {\paren {q \land r} \implies q} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} }$ | Axiom $A 4$ | $\paren {q \land r} \, / \, q, q \, / \, r$ | |||
3 | $\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q}$ | Rule $RST \, 3$ | 1, 2 | |||
4 | $\paren {q \land r} \implies \paren {r \land q}$ | Rule of Commutation | $q / p, r / q$ | |||
5 | $\paren {r \land q} \implies r$ | Rule of Simplification | $r / p$ | |||
6 | $\paren {q \land r} \implies r$ | Hypothetical Syllogism | 4, 5 | |||
7 | $\paren {\paren {q \land r} \implies r} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor r} }$ | Axiom $A4$ | $\paren {q \land r} \, / \, q$ | |||
8 | $\paren {p \lor \paren {q \land r} } \implies \paren {p \lor r}$ | Rule $RST \, 3$ | 6, 7 | |||
9 | $\paren {p \lor q} \implies \paren {\paren {p \lor r} \implies \paren {\paren {p \lor q} \land \paren {p \lor r} } }$ | Rule of Conjunction | $\paren {p \lor q} \, / \, p, \paren {p \lor r} \, / \, q$ | |||
10 | $\paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor r} \implies \paren {\paren {p \lor q} \land \paren {p \lor r} } }$ | Hypothetical Syllogism | 3, 9 | |||
11 | $\paren {\paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor r} \implies \paren {p \lor q} \land \paren {p \lor r} } } \\ \implies \paren {\paren {p \lor r} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} } }$ | Principle of Commutation | $\paren {p \lor \paren {q \land r} } \, / \, p, \paren {p \lor r} \, / \, q, \paren {\paren {p \lor q} \land \paren {p \lor r} } \, / \, r$ | |||
12 | $\paren {p \lor r} \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} }$ | Rule $RST \, 3$ | 10, 11 | |||
13 | $\paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r} }$ | Hypothetical Syllogism | 8, 12 | |||
14 | $\paren {p \lor \paren {q \land r} } \implies \paren {p \lor q} \land \paren {p \lor r}$ |
$\blacksquare$