29 lines
682 B
Text
29 lines
682 B
Text
theorem ex1 (p q r : Prop) (h1 : p ∨ q) (h2 : p → q) : q :=
|
||
have : q := by -- Error here
|
||
skip
|
||
by skip -- Error here
|
||
skip
|
||
|
||
theorem ex2 (p q r : Prop) (h1 : p ∨ q) (h2 : p → q) : q :=
|
||
have : q := by {
|
||
skip
|
||
} -- Error here
|
||
by skip -- Error here
|
||
skip
|
||
|
||
theorem ex3 (p q r : Prop) (h1 : p ∨ q) (h2 : p → q) : q := by
|
||
cases h1
|
||
{ skip
|
||
skip } -- Error here
|
||
{ skip
|
||
skip } -- Error here
|
||
|
||
theorem ex4 (p q r : Prop) (h1 : p ∨ q) (h2 : p → q) : q := by
|
||
first | done | apply ex3 p q r h1 h2
|
||
|
||
theorem ex5 (p q r : Prop) (h1 : p ∨ q) (h2 : p → q) : q := by
|
||
cases h1
|
||
. skip -- Error here
|
||
skip
|
||
. skip -- Error here
|
||
skip
|