[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 5 from proof inrec (
previous step
)
a × d ∈ ax0, ∀ a0, d2, b0, d3. a0 × d2 ∈ ax0 ∨ a0 = b0 → (∃ d1. a0 × d1 ∈ ax0 ∨ a0 = b0 ∧ d1 = d3) ⊦ ∃ d1. a × d1 = b × d0 ∨ a × d1 = p ∨ a × d1 ∈ ax0
Heuristic quantifier closing applied rule exists right
And closed the goal!
Rule argument
pos: right 1 terms: d
Additional information
Used terms: d