[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 1 from proof bitsand-zeros (conclusion)
⊦ ¬ true ∈ bits ∧ (# bits ≤ # bits0 ∨ bits0.first) ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bits
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
# bits ≤ # bits0 ∨ bits0.first ⊦ bits = [], bits0 = [], bitsand(bits, bits0) = bits, true ∈ bits
Rule argument
Additional information