[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 eq-01 (
previous step
)
a ≠ a → (a ∈ s ↔ a ∈ s0), ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a0 ∈ s ↔ a0 ∈ s0, a ∈ s0, a ∈ s
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a0 ∈ s ↔ a0 ∈ s0, a ∈ s0, a ∈ s
Rule argument
Additional information