[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 links-dirs-union (
previous step
)
dirs0 ⊥ dirs1 ⊦ links(ino, dirs0 ∪ dirs1) = links(ino, dirs0) ∪ links(ino, dirs1)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
dirs0 ⊥ dirs1 ⊦ links(ino, dirs0 ∪ dirs1) = links(ino, dirs0) ∪ links(ino, dirs1)
Rule argument
Additional information