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Step * of Lemma insert-int-comm

∀T:Type. ∀a,b:T.  ((T ⊆r ℤ) ⇒ (∀L:T List. (insert-int(b;insert-int(a;L)) = insert-int(a;insert-int(b;L)) ∈ (T List))))
BY
{ Assert ⌜∀L:ℤ List. ∀a,b:ℤ.  (insert-int(b;insert-int(a;L)) = insert-int(a;insert-int(b;L)) ∈ (ℤ List))⌝⋅ }

1
.....assertion..... 
∀L:ℤ List. ∀a,b:ℤ.  (insert-int(b;insert-int(a;L)) = insert-int(a;insert-int(b;L)) ∈ (ℤ List))

2
1. ∀L:ℤ List. ∀a,b:ℤ.  (insert-int(b;insert-int(a;L)) = insert-int(a;insert-int(b;L)) ∈ (ℤ List))
⊢ ∀T:Type. ∀a,b:T.
    ((T ⊆r ℤ) ⇒ (∀L:T List. (insert-int(b;insert-int(a;L)) = insert-int(a;insert-int(b;L)) ∈ (T List))))

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Latex:

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Latex:
\mforall{}T:Type.  \mforall{}a,b:T.
    ((T  \msubseteq{}r  \mBbbZ{})  {}\mRightarrow{}  (\mforall{}L:T  List.  (insert-int(b;insert-int(a;L))  =  insert-int(a;insert-int(b;L)))))


By

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Latex:
Assert  \mkleeneopen{}\mforall{}L:\mBbbZ{}  List.  \mforall{}a,b:\mBbbZ{}.    (insert-int(b;insert-int(a;L))  =  insert-int(a;insert-int(b;L)))\mkleeneclose{}\mcdot{}

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