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mardurhack

ex1 matita

Jun 9th, 2011
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  1. theorem ex1 : (A ⇒ C) ⇒ (B ⇒ C) ⇒ A ∨ B ⇒ C.
  2. apply rule (prove ((A ⇒ C) ⇒ (B ⇒ C) ⇒ A ∨ B ⇒ C));
  3. apply rule (⇒#i [H1] ((B⇒ C) ⇒ A ∨ B ⇒ C));
  4. apply rule (⇒#i [H2] (A ∨ B ⇒ C));
  5. apply rule (⇒#i [H3] (C));
  6. apply rule (∨#e (A ∨ B) [H01] (C) [H02] (C));
  7. [apply rule (discharge [H3]);
  8. |apply rule (⇒#e (A ⇒ C) (A));
  9. [apply rule (discharge [H1]);
  10. |apply rule (discharge [H01]);
  11. ]
  12. |apply rule (⇒#e (B ⇒ C) (B));
  13. [apply rule (discharge [H2]);
  14. |apply rule (discharge [H02]);
  15. ]
  16. ]
  17.  
  18. qed.
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