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Past exam of the logic and set theory course of the University of Oxford
/
2007
/
B
/
5
/
iii
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(
@codex,
0
)
...
Year 3 of the mathematics course of the University of Oxford
Logic and set theory course of the University of Oxford
Past exam of the logic and set theory course of the University of Oxford
2007
B
5
2026-10-31
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Solution
iii
Solution
0
0
0
iii
Fix
m
and induct on
n
. At
n
=
0
,
m
+
+
0
=
m
+
=
(
m
+
0
)
+
=
m
+
0
+
.
(1)
If
m
+
+
n
=
m
+
n
+
, then
m
+
+
n
+
=
(
m
+
+
n
)
+
=
(
m
+
n
+
)
+
=
m
+
(
n
+
)
+
.
(2)
The last expression is the right-hand side required at the successor
n
+
. Therefore
m
+
+
n
=
m
+
n
+
(
m
,
n
∈
ω
)
.
(3)
In particular both sides also equal
(
m
+
n
)
+
by the
recursion
equation
for the
second
argument.
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(12)
5
B
2007
Past exam of the logic and set theory course of the University of Oxford
Logic and set theory course of the University of Oxford
Year 3 of the mathematics course of the University of Oxford
Mathematics course of the University of Oxford
Undergraduate course of the University of Oxford
Course of the University of Oxford
University of Oxford
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