[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 2 from proof subset-plus (previous step)

bag0 ++ a ⊆ bag1 ⊦ ∃ bag2. bag1 = bag2 ++ a ∧ bag0 ⊆ bag2

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ bag → (∃ bag0. bag = bag0 ++ a)
Proof step 4:
bag0 ++ a ⊆ bag1, a ∈ bag1 → (∃ bag0. bag1 = bag0 ++ a) ⊦ ∃ bag2. bag1 = bag2 ++ a ∧ bag0 ⊆ bag2

Rule argument

spec: <toplevel> name: exin seq: ⊦ a ∈ bag → (∃ bag0. bag = bag0 ++ a) subst: [a, bag] → [a, bag1]

Additional information

Used terms: a, bag1