Step
*
1
of Lemma
agree_on_common_append
1. [T] : Type
⊢ ∀bs,cs,ds:T List.
(agree_on_common(T;[];cs)
⇒ agree_on_common(T;bs;ds)
⇒ agree_on_common(T;bs;cs @ ds)) supposing
((∀x∈[].¬(x ∈ ds)) and
(∀x∈cs.¬(x ∈ bs)))
BY
{ ((((D 0 THENA Auto) THEN InductionOnList) THEN Reduce 0) THEN Auto) }
1
1. [T] : Type
2. bs : T List
3. u : T
4. v : T List
5. ∀ds:T List
(agree_on_common(T;[];v)
⇒ agree_on_common(T;bs;ds)
⇒ agree_on_common(T;bs;v @ ds)) supposing
((∀x∈[].¬(x ∈ ds)) and
(∀x∈v.¬(x ∈ bs)))
6. ds : T List
7. (∀x∈[u / v].¬(x ∈ bs))
8. (∀x∈[].¬(x ∈ ds))
9. agree_on_common(T;[];[u / v])
10. agree_on_common(T;bs;ds)
⊢ agree_on_common(T;bs;[u / (v @ ds)])
Latex:
Latex:
1. [T] : Type
\mvdash{} \mforall{}bs,cs,ds:T List.
(agree\_on\_common(T;[];cs)
{}\mRightarrow{} agree\_on\_common(T;bs;ds)
{}\mRightarrow{} agree\_on\_common(T;bs;cs @ ds)) supposing
((\mforall{}x\mmember{}[].\mneg{}(x \mmember{} ds)) and
(\mforall{}x\mmember{}cs.\mneg{}(x \mmember{} bs)))
By
Latex:
((((D 0 THENA Auto) THEN InductionOnList) THEN Reduce 0) THEN Auto)
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