Hypothetical Syllogism/Formulation 4/Proof 2
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Theorem
- $\vdash \paren {p \implies q} \implies \paren {\paren {q \implies r} \implies \paren {p \implies r} }$
Proof
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
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 3.4$: Theorem $3.16$