Step
*
of Lemma
type-separation
∀x,y:Base.
(((x)↓ ∨ is-exception(x))
⇒ ((y)↓ ∨ is-exception(y))
⇒ (∀n,m:ℤ. ∀T:Type. ((x = n ∈ T)
⇒ (y = m ∈ T)
⇒ (x = y ∈ T)))
⇒ (x = y ∈ Base))
BY
{ (Auto THEN Assert ⌜∃n,m:ℤ. ((¬(n = m ∈ ℤ)) ∧ (¬(x = m ∈ Base)) ∧ (¬(y = n ∈ Base)))⌝⋅) }
1
.....assertion.....
1. x : Base
2. y : Base
3. (x)↓ ∨ is-exception(x)
4. (y)↓ ∨ is-exception(y)
5. ∀n,m:ℤ. ∀T:Type. ((x = n ∈ T)
⇒ (y = m ∈ T)
⇒ (x = y ∈ T))
⊢ ∃n,m:ℤ. ((¬(n = m ∈ ℤ)) ∧ (¬(x = m ∈ Base)) ∧ (¬(y = n ∈ Base)))
2
1. x : Base
2. y : Base
3. (x)↓ ∨ is-exception(x)
4. (y)↓ ∨ is-exception(y)
5. ∀n,m:ℤ. ∀T:Type. ((x = n ∈ T)
⇒ (y = m ∈ T)
⇒ (x = y ∈ T))
6. ∃n,m:ℤ. ((¬(n = m ∈ ℤ)) ∧ (¬(x = m ∈ Base)) ∧ (¬(y = n ∈ Base)))
⊢ x = y ∈ Base
Latex:
Latex:
\mforall{}x,y:Base.
(((x)\mdownarrow{} \mvee{} is-exception(x))
{}\mRightarrow{} ((y)\mdownarrow{} \mvee{} is-exception(y))
{}\mRightarrow{} (\mforall{}n,m:\mBbbZ{}. \mforall{}T:Type. ((x = n) {}\mRightarrow{} (y = m) {}\mRightarrow{} (x = y)))
{}\mRightarrow{} (x = y))
By
Latex:
(Auto THEN Assert \mkleeneopen{}\mexists{}n,m:\mBbbZ{}. ((\mneg{}(n = m)) \mwedge{} (\mneg{}(x = m)) \mwedge{} (\mneg{}(y = n)))\mkleeneclose{}\mcdot{})
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