Hypothetical Syllogism/Formulation 4
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Theorem
- $\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
Proof 1
Let us use the following abbreviations
\(\ds \phi\) | \(\text{ for }\) | \(\ds p \implies q\) | ||||||||||||
\(\ds \psi\) | \(\text{ for }\) | \(\ds q \implies r\) | ||||||||||||
\(\ds \chi\) | \(\text{ for }\) | \(\ds p \implies r\) |
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | $\paren {\phi \land \psi} \implies \chi$ | Theorem Introduction | (None) | Hypothetical Syllogism: Formulation 3 | ||
2 | $\phi \implies \paren {\psi \implies \chi}$ | Sequent Introduction | 1 | Rule of Exportation |
Expanding the abbreviations leads us back to:
- $\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
$\blacksquare$
Proof 2
This proof is derived in the context of the following proof system: instance 1 of a Hilbert proof system.
By the tableau method:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Assumption | (None) | ||
2 | 2 | $p \implies q$ | Assumption | (None) | ||
3 | 1, 2 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 2 | ||
4 | 4 | $q \implies r$ | Assumption | (None) | ||
5 | 1, 2, 4 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 3, 4 | ||
6 | 2, 4 | $p \implies r$ | Deduction Rule | 5 | ||
7 | 2 | $\paren {q \implies r} \implies \paren {p \implies r}$ | Deduction Rule | 6 | ||
8 | $\paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$ | Deduction Rule | 7 |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T4}$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(2) \ \text{(iii)}$