[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 look-del (
previous step
)
∀ r0, sk. r ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r) ∧ st[r0 -- sk] = (st -- r)[r0 -- sk] ⊦ (st -- r)[r0 -- sk0] = st[r0 -- sk0], r = r0, r ∈ st -- r
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
[EXRARG : pos: left 1, [QUANTTERMLIST : (r0, sk0)]]
Additional information
[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r0, sk0)]]