Metatheory - University of Cambridge

5. Soundness 40
m
i
j
k
l
A ∨ B
A
C
B
C
n C ∨E m, i–j, k–l
Let v be any valuation that makes all of ∆ n true. Note that all of ∆ m are
among ∆ n . By hypothesis, line m is shiny. So any valuation that makes ∆ n
true makes A ∨ B true. So in particular, v makes A ∨ B true, and hence either
v makes A true, or v makes B true. We now reason through these two cases:
Case 1: v makes A true. All of ∆ i are among ∆ n , with the possible exception
of A. Since v makes all of ∆ n true, and also makes A true, v
makes all of ∆ i true. Now, by assumption, line j is shiny; so ∆ j ⊨ C . But
the sentences ∆ i are just the sentences ∆ j , so ∆ i ⊨ C . So, any valuation
that makes all of ∆ i true makes C true. But v is just such a valuation.
So v makes C true.
Case 2: v makes B true. Reasoning in exactly the same way, considering
lines k and l, v makes C true.
Either way, v makes C true. So ∆ n ⊨ C .
■
Lemma 5.8. ⊥I is rule-sound.
Proof. Assume that every line before line n on some TFL-proof is shiny, and
that ⊥I is used on line n. So the situation is:
i
A
j ¬A
n ⊥ ⊥I i, j
Note that all of ∆ i and all of ∆ j are among ∆ n . By hypothesis, lines i and j
are shiny. So any valuation which makes all of ∆ n true would have to make
both A and ¬A true. But no valuation can do that. So no valuation makes all
of ∆ n true. So ∆ n ⊨ ⊥, vacuously.
■
Lemma 5.9. ⊥E is rule-sound.
Proof. I leave this as an exercise.
■
Lemma 5.10. ¬I is rule-sound.
Proof. Assume that every line before line n on some TFL-proof is shiny, and
that ¬I is used on line n. So the situation is: