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proc.py

@@ -10,5 +10,7 @@ for i, row in df.iterrows():
         df.drop(i)
         continue
     split = stmt.split('\n')
-    stmt = "\n".join([i for i in split if not i.startswith("abbrev")])
+    stmt = "\n".join([i.replace("sorry", "") for i in split if not i.startswith("abbrev")])
+    print(stmt)
+    print("*"*50)
 df.to_csv('putnam.csv', index=False)

putnam.csv

@@ -4261,7 +4261,8 @@ putnam_2017_b4,"abbrev putnam_2017_b4_solution : ℝ := sorry
 -- (log 2) ^ 2
 theorem putnam_2017_b4
 : (∑' k : ℕ, 3 * log (4 * k + 2) / (4 * k + 2) - log (4 * k + 3) / (4 * k + 3) - log (4 * k + 4) / (4 * k + 4) - log (4 * k + 5) / (4 * k + 5) = putnam_2017_b4_solution) :=
-sorry","Evaluate the sum \begin{gather*} \sum_{k=0}^\infty \left( 3 \cdot \frac{\ln(4k+2)}{4k+2} - \frac{\ln(4k+3)}{4k+3} - \frac{\ln(4k+4)}{4k+4} - \frac{\ln(4k+5)}{4k+5} 
+sorry","Evaluate the sum \begin{gather*} \sum_{k=0}^\infty \left( 3 \cdot \frac{\ln(4k+2)}{4k+2} - \frac{\ln(4k+3)}{4k+3} - \frac{\ln(4k+4)}{4k+4} - \frac{\ln(4k+5)}{4k+5} 
+ight) \ = 3 \cdot \frac{\ln 2}{2} - \frac{\ln 3}{3} - \frac{\ln 4}{4} - \frac{\ln 5}{5} + 3 \cdot \frac{\ln 6}{6} - \frac{\ln 7}{7} \ - \frac{\ln 8}{8} - \frac{\ln 9}{9} + 3 \cdot \frac{\ln 10}{10} - \cdots . \end{gather*} (As usual, $\ln x$ denotes the natural logarithm of $x$.)",Prove that the sum equals $(\ln 2)^2$.,['algebra'],"Section putnam_2017_b4.
 Require Import Reals Coquelicot.Coquelicot.
 Definition putnam_2017_b4_solution := (ln 2 / ln 10) ^ 2.
 Theorem putnam_2017_b4