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l |
Need to show that
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p1',
…, pn', (p1 … pn q) ╞ qθ
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provided that pi'θ = piθ
for all I
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Lemma: For any
sentence p, we have p ╞ pθ by UI
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1. |
(p1
… pn q) ╞ (p1
… pn q)θ = (p1θ
… pnθ
qθ)
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2. |
p1',
…, pn' ╞ p1' … pn' ╞ p1'θ … pn'θ
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3. |
From 1 and 2, qθ
follows by ordinary Modus Ponens
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