diff options
Diffstat (limited to 'test')
| -rw-r--r-- | test/phase5/exact-proof-parity-invalid-assume.tex | 8 | ||||
| -rw-r--r-- | test/phase5/exact-proof-parity-invalid-fix-shape.tex | 8 | ||||
| -rw-r--r-- | test/phase5/exact-proof-parity-invalid-fix.tex | 8 | ||||
| -rw-r--r-- | test/phase5/exact-proof-parity.tex | 107 |
4 files changed, 131 insertions, 0 deletions
diff --git a/test/phase5/exact-proof-parity-invalid-assume.tex b/test/phase5/exact-proof-parity-invalid-assume.tex new file mode 100644 index 0000000..3bb04bc --- /dev/null +++ b/test/phase5/exact-proof-parity-invalid-assume.tex @@ -0,0 +1,8 @@ +\begin{proposition}\label{invalid_disjunct_assume} + Let $A,B$ be sets. + $A=A$ or $B=B$. +\end{proposition} +\begin{proof} + Assume $A=A$. + Follows. +\end{proof} diff --git a/test/phase5/exact-proof-parity-invalid-fix-shape.tex b/test/phase5/exact-proof-parity-invalid-fix-shape.tex new file mode 100644 index 0000000..16a58a2 --- /dev/null +++ b/test/phase5/exact-proof-parity-invalid-fix-shape.tex @@ -0,0 +1,8 @@ +\begin{proposition}\label{invalid_fix_shape} + Let $A$ be a set. + $A=A$. +\end{proposition} +\begin{proof} + Fix $x\in A$. + Follows. +\end{proof} diff --git a/test/phase5/exact-proof-parity-invalid-fix.tex b/test/phase5/exact-proof-parity-invalid-fix.tex new file mode 100644 index 0000000..0a2b24c --- /dev/null +++ b/test/phase5/exact-proof-parity-invalid-fix.tex @@ -0,0 +1,8 @@ +\begin{proposition}\label{invalid_bounded_fix} + Let $A$ be a set. + For all $x\in A$ we have $x\in A$. +\end{proposition} +\begin{proof} + Fix $x\notin A$. + Follows. +\end{proof} diff --git a/test/phase5/exact-proof-parity.tex b/test/phase5/exact-proof-parity.tex new file mode 100644 index 0000000..73d849f --- /dev/null +++ b/test/phase5/exact-proof-parity.tex @@ -0,0 +1,107 @@ +\begin{proposition}\label{bounded_fix_single} + Let $A$ be a set. + For all $x\in A$ we have $x\in A$. +\end{proposition} +\begin{proof} + Fix $x\in A$. + Follows by assumption. +\end{proof} + +\begin{proposition}\label{bounded_fix_multiple} + Let $A$ be a set. + For all $x,y\in A$ we have $x\in A$ and $y\in A$. +\end{proposition} +\begin{proof} + Fix $x,y\in A$. + Follows by assumption. +\end{proof} + +\begin{proposition}\label{bounded_fix_negative} + Let $A$ be a set. + For all $x\notin A$ we have $x\notin A$. +\end{proposition} +\begin{proof} + Fix $x\notin A$. + Follows by assumption. +\end{proof} + +\begin{proposition}\label{fix_such_that} + Let $A$ be a set. + For all $x$ such that $x\in A$ we have $x\in A$. +\end{proposition} +\begin{proof} + Fix $x$ such that $x\in A$. + Follows by assumption. +\end{proof} + +\begin{proposition}\label{assume_left_conjunct} + Let $A,B$ be sets. + If $A=A$ and $B=B$, then $A=A$. +\end{proposition} +\begin{proof} + Assume $A=A$. + Assume $B=B$. + Follows by assumption. +\end{proof} + +\begin{proposition}\label{assume_right_conjunct} + Let $A,B$ be sets. + If $A=A$ and $B=B$, then $B=B$. +\end{proposition} +\begin{proof} + Assume $B=B$. + Assume $A=A$. + Follows by assumption. +\end{proof} + +\begin{proposition}\label{take_bounded} + Let $A$ be a set. + Suppose there exists $x\in A$ such that $x=x$. + Then $A=A$. +\end{proposition} +\begin{proof} + Take $x\in A$ such that $x=x$ by assumption. + We have $x\in A$ by assumption. + Follows. +\end{proof} + +\begin{proposition}\label{take_named_noun} + Let $A$ be a set. + Suppose there exist sets $x,y$ such that $x=x$ and $y=y$. + Then $A=A$. +\end{proposition} +\begin{proof} + Take a set $x,y$ such that $x=x$ and $y=y$ by assumption. + Follows. +\end{proof} + +\begin{proposition}\label{take_anonymous_noun} + Let $A$ be a set. + Suppose there exists a set. + Then $A=A$. +\end{proposition} +\begin{proof} + Take a set by assumption. + Follows. +\end{proof} + +\begin{proposition}\label{existential_have_witness} + Let $A$ be a set. + Suppose there exists $x\in A$ such that $x=x$. + Then $A=A$. +\end{proposition} +\begin{proof} + We have there exists $x\in A$ such that $x=x$ by assumption. + We have $x\in A$ by assumption. + Follows. +\end{proof} + +\begin{proposition}\label{take_omitted_continuation} + Let $A$ be a set. + Suppose there exists $x\in A$ such that $x=x$. + Then $A=A$. +\end{proposition} +\begin{proof} + Take $x\in A$ such that $x=x$ by assumption. + Omitted. +\end{proof} |
