[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 intersection-shift (
previous step
)
a ∈ s, a1 ∈ s, a0 = a ∨ a0 ∈ s ∧ a0 ∈ s0 ⊦ a0 ∈ s ∧ (a0 = a ∨ a0 ∈ s0)
User applied rule case distinction
And closed the goal!
Rule argument
pos: left 3
Additional information