Two automata, one parity languageComplete reachable product · 3 state pairs
The problem
Decide whether two complete deterministic automata accept the same binary strings.
The checked result
Equivalent: yes.
Starting from the initial pair, the checker explores every reachable pair of states and requires matching acceptance. No mismatch is reachable, so equivalence holds for every finite binary word. Both accept an even number of ones.
Why the checker accepts it
- Start from the pair of initial states.
- Explore every symbol transition until no new pair is reachable.
- Compare acceptance in every reachable pair.
- For inequivalence, replay a distinguishing input from both initial states.
Formal specification
{
"alphabet": [
"0",
"1"
],
"left": {
"start": "E",
"accepting": [
"E"
],
"transitions": {
"E": {
"0": "E",
"1": "O"
},
"O": {
"0": "O",
"1": "E"
}
}
}
}Claim and evidence
{
"claim": {
"equivalent": true
},
"witness": {
"right": {
"start": "A",
"accepting": [
"A",
"B"
],
"transitions": {
"A": {
"0": "B",
"1": "C"
},
"B": {
"0": "A",
"1": "C"
},
"C": {
"0": "C",
"1": "A"
}
}
}
}
}Dataset construction
Deterministic finite fixture; full enumeration or witness replay as stated. Structured JSON; field meanings are stated in the specification. Acceptance covers 3 checker units for this record; the unit type is stated in its verification scope.
Complexity and limits
At most |Q₁|·|Q₂| product states are visited, with one outgoing edge per alphabet symbol.
Applies only to these two specified complete deterministic automata; no claim of automaton minimality.
A boundary to investigate
Bounded string testing is weaker than complete DFA product exploration. Extend the checker to emit shortest distinguishing words and state-minimization partitions.