304 étapes
Wiki de communauté, car cette preuve est générée par la fonction FindEquationalProof de Mathematica .
La preuve est assez longue. Mathematica ne sait pas comment jouer au golf.
Ceci est le code Mathematica qui génère la preuve (nécessite Mathematica 11.3), où p, t, des nmoyens +, ×, -respectivement:
ringAxioms = {ForAll[{a, b, c}, p[a, p[b, c]] == p[p[a, b], c]],
ForAll[a, p[a, 0] == a],
ForAll[a, p[a, n[a]] == 0],
ForAll[{a, b}, p[a, b] == p[b, a]],
ForAll[{a, b, c}, t[a, t[b, c]] == t[t[a, b], c]],
ForAll[a, t[a, 1] == a], ForAll[a, t[1, a] == a],
ForAll[{a, b, c}, t[a, p[b, c]] == p[t[a, b], t[a, c]]],
ForAll[{a, b, c}, t[p[b, c], a] == p[t[b, a], t[c, a]]]};
proof = FindEquationalProof[t[n[a], n[a]] == t[a, a], ringAxioms];
proof["ProofNotebook"]
Il n’est pas facile de compter les pas directement, je le calcule donc par le nombre de chemins allant des axiomes à la conclusion dans le "graphe de preuve".
graph = proof["ProofGraph"];
score = Sum[
Length[FindPath[graph, axiom, "Conclusion 1", Infinity,
All]], {axiom,
Select[VertexList[graph], StringMatchQ["Axiom " ~~ __]]}]
Essayez-le en ligne!
C'est la preuve générée par le code:
Axiom 1
We are given that:
x1==p[x1, 0]
Axiom 2
We are given that:
x1==t[x1, 1]
Axiom 3
We are given that:
x1==t[1, x1]
Axiom 4
We are given that:
p[x1, x2]==p[x2, x1]
Axiom 5
We are given that:
p[x1, p[x2, x3]]==p[p[x1, x2], x3]
Axiom 6
We are given that:
p[x1, n[x1]]==0
Axiom 7
We are given that:
p[t[x1, x2], t[x3, x2]]==t[p[x1, x3], x2]
Axiom 8
We are given that:
p[t[x1, x2], t[x1, x3]]==t[x1, p[x2, x3]]
Axiom 9
We are given that:
t[x1, t[x2, x3]]==t[t[x1, x2], x3]
Hypothesis 1
We would like to show that:
t[n[a], n[a]]==t[a, a]
Critical Pair Lemma 1
The following expressions are equivalent:
p[0, x1]==x1
Proof
Note that the input for the rule:
p[x1_, x2_]\[TwoWayRule]p[x2_, x1_]
contains a subpattern of the form:
p[x1_, x2_]
which can be unified with the input for the rule:
p[x1_, 0]->x1
where these rules follow from Axiom 4 and Axiom 1 respectively.
Critical Pair Lemma 2
The following expressions are equivalent:
p[x1, p[n[x1], x2]]==p[0, x2]
Proof
Note that the input for the rule:
p[p[x1_, x2_], x3_]->p[x1, p[x2, x3]]
contains a subpattern of the form:
p[x1_, x2_]
which can be unified with the input for the rule:
p[x1_, n[x1_]]->0
where these rules follow from Axiom 5 and Axiom 6 respectively.
Critical Pair Lemma 3
The following expressions are equivalent:
t[p[1, x1], x2]==p[x2, t[x1, x2]]
Proof
Note that the input for the rule:
p[t[x1_, x2_], t[x3_, x2_]]->t[p[x1, x3], x2]
contains a subpattern of the form:
t[x1_, x2_]
which can be unified with the input for the rule:
t[1, x1_]->x1
where these rules follow from Axiom 7 and Axiom 3 respectively.
Critical Pair Lemma 4
The following expressions are equivalent:
t[x1, p[1, x2]]==p[x1, t[x1, x2]]
Proof
Note that the input for the rule:
p[t[x1_, x2_], t[x1_, x3_]]->t[x1, p[x2, x3]]
contains a subpattern of the form:
t[x1_, x2_]
which can be unified with the input for the rule:
t[x1_, 1]->x1
where these rules follow from Axiom 8 and Axiom 2 respectively.
Critical Pair Lemma 5
The following expressions are equivalent:
t[p[1, x1], 0]==t[x1, 0]
Proof
Note that the input for the rule:
p[x1_, t[x2_, x1_]]->t[p[1, x2], x1]
contains a subpattern of the form:
p[x1_, t[x2_, x1_]]
which can be unified with the input for the rule:
p[0, x1_]->x1
where these rules follow from Critical Pair Lemma 3 and Critical Pair Lemma 1 respectively.
Critical Pair Lemma 6
The following expressions are equivalent:
t[0, 0]==t[1, 0]
Proof
Note that the input for the rule:
t[p[1, x1_], 0]->t[x1, 0]
contains a subpattern of the form:
p[1, x1_]
which can be unified with the input for the rule:
p[x1_, 0]->x1
where these rules follow from Critical Pair Lemma 5 and Axiom 1 respectively.
Substitution Lemma 1
It can be shown that:
t[0, 0]==0
Proof
We start by taking Critical Pair Lemma 6, and apply the substitution:
t[1, x1_]->x1
which follows from Axiom 3.
Critical Pair Lemma 7
The following expressions are equivalent:
t[x1, 0]==t[p[x1, 1], 0]
Proof
Note that the input for the rule:
t[p[1, x1_], 0]->t[x1, 0]
contains a subpattern of the form:
p[1, x1_]
which can be unified with the input for the rule:
p[x1_, x2_]\[TwoWayRule]p[x2_, x1_]
where these rules follow from Critical Pair Lemma 5 and Axiom 4 respectively.
Critical Pair Lemma 8
The following expressions are equivalent:
t[0, p[1, x1]]==t[0, x1]
Proof
Note that the input for the rule:
p[x1_, t[x1_, x2_]]->t[x1, p[1, x2]]
contains a subpattern of the form:
p[x1_, t[x1_, x2_]]
which can be unified with the input for the rule:
p[0, x1_]->x1
where these rules follow from Critical Pair Lemma 4 and Critical Pair Lemma 1 respectively.
Critical Pair Lemma 9
The following expressions are equivalent:
t[p[x1, 1], p[1, 0]]==p[p[x1, 1], t[x1, 0]]
Proof
Note that the input for the rule:
p[x1_, t[x1_, x2_]]->t[x1, p[1, x2]]
contains a subpattern of the form:
t[x1_, x2_]
which can be unified with the input for the rule:
t[p[x1_, 1], 0]->t[x1, 0]
where these rules follow from Critical Pair Lemma 4 and Critical Pair Lemma 7 respectively.
Substitution Lemma 2
It can be shown that:
t[p[x1, 1], 1]==p[p[x1, 1], t[x1, 0]]
Proof
We start by taking Critical Pair Lemma 9, and apply the substitution:
p[x1_, 0]->x1
which follows from Axiom 1.
Substitution Lemma 3
It can be shown that:
p[x1, 1]==p[p[x1, 1], t[x1, 0]]
Proof
We start by taking Substitution Lemma 2, and apply the substitution:
t[x1_, 1]->x1
which follows from Axiom 2.
Substitution Lemma 4
It can be shown that:
p[x1, 1]==p[x1, p[1, t[x1, 0]]]
Proof
We start by taking Substitution Lemma 3, and apply the substitution:
p[p[x1_, x2_], x3_]->p[x1, p[x2, x3]]
which follows from Axiom 5.
Critical Pair Lemma 10
The following expressions are equivalent:
t[0, x1]==t[0, p[x1, 1]]
Proof
Note that the input for the rule:
t[0, p[1, x1_]]->t[0, x1]
contains a subpattern of the form:
p[1, x1_]
which can be unified with the input for the rule:
p[x1_, x2_]\[TwoWayRule]p[x2_, x1_]
where these rules follow from Critical Pair Lemma 8 and Axiom 4 respectively.
Critical Pair Lemma 11
The following expressions are equivalent:
t[p[1, 0], p[x1, 1]]==p[p[x1, 1], t[0, x1]]
Proof
Note that the input for the rule:
p[x1_, t[x2_, x1_]]->t[p[1, x2], x1]
contains a subpattern of the form:
t[x2_, x1_]
which can be unified with the input for the rule:
t[0, p[x1_, 1]]->t[0, x1]
where these rules follow from Critical Pair Lemma 3 and Critical Pair Lemma 10 respectively.
Substitution Lemma 5
It can be shown that:
t[1, p[x1, 1]]==p[p[x1, 1], t[0, x1]]
Proof
We start by taking Critical Pair Lemma 11, and apply the substitution:
p[x1_, 0]->x1
which follows from Axiom 1.
Substitution Lemma 6
It can be shown that:
p[x1, 1]==p[p[x1, 1], t[0, x1]]
Proof
We start by taking Substitution Lemma 5, and apply the substitution:
t[1, x1_]->x1
which follows from Axiom 3.
Substitution Lemma 7
It can be shown that:
p[x1, 1]==p[x1, p[1, t[0, x1]]]
Proof
We start by taking Substitution Lemma 6, and apply the substitution:
p[p[x1_, x2_], x3_]->p[x1, p[x2, x3]]
which follows from Axiom 5.
Substitution Lemma 8
It can be shown that:
p[x1, p[n[x1], x2]]==x2
Proof
We start by taking Critical Pair Lemma 2, and apply the substitution:
p[0, x1_]->x1
which follows from Critical Pair Lemma 1.
Critical Pair Lemma 12
The following expressions are equivalent:
n[n[x1]]==p[x1, 0]
Proof
Note that the input for the rule:
p[x1_, p[n[x1_], x2_]]->x2
contains a subpattern of the form:
p[n[x1_], x2_]
which can be unified with the input for the rule:
p[x1_, n[x1_]]->0
where these rules follow from Substitution Lemma 8 and Axiom 6 respectively.
Substitution Lemma 9
It can be shown that:
n[n[x1]]==x1
Proof
We start by taking Critical Pair Lemma 12, and apply the substitution:
p[x1_, 0]->x1
which follows from Axiom 1.
Critical Pair Lemma 13
The following expressions are equivalent:
x1==p[n[x2], p[x2, x1]]
Proof
Note that the input for the rule:
p[x1_, p[n[x1_], x2_]]->x2
contains a subpattern of the form:
n[x1_]
which can be unified with the input for the rule:
n[n[x1_]]->x1
where these rules follow from Substitution Lemma 8 and Substitution Lemma 9 respectively.
Critical Pair Lemma 14
The following expressions are equivalent:
t[x1, x2]==p[n[x2], t[p[1, x1], x2]]
Proof
Note that the input for the rule:
p[n[x1_], p[x1_, x2_]]->x2
contains a subpattern of the form:
p[x1_, x2_]
which can be unified with the input for the rule:
p[x1_, t[x2_, x1_]]->t[p[1, x2], x1]
where these rules follow from Critical Pair Lemma 13 and Critical Pair Lemma 3 respectively.
Critical Pair Lemma 15
The following expressions are equivalent:
t[x1, x2]==p[n[x1], t[x1, p[1, x2]]]
Proof
Note that the input for the rule:
p[n[x1_], p[x1_, x2_]]->x2
contains a subpattern of the form:
p[x1_, x2_]
which can be unified with the input for the rule:
p[x1_, t[x1_, x2_]]->t[x1, p[1, x2]]
where these rules follow from Critical Pair Lemma 13 and Critical Pair Lemma 4 respectively.
Critical Pair Lemma 16
The following expressions are equivalent:
p[1, t[x1, 0]]==p[n[x1], p[x1, 1]]
Proof
Note that the input for the rule:
p[n[x1_], p[x1_, x2_]]->x2
contains a subpattern of the form:
p[x1_, x2_]
which can be unified with the input for the rule:
p[x1_, p[1, t[x1_, 0]]]->p[x1, 1]
where these rules follow from Critical Pair Lemma 13 and Substitution Lemma 4 respectively.
Substitution Lemma 10
It can be shown that:
p[1, t[x1, 0]]==1
Proof
We start by taking Critical Pair Lemma 16, and apply the substitution:
p[n[x1_], p[x1_, x2_]]->x2
which follows from Critical Pair Lemma 13.
Critical Pair Lemma 17
The following expressions are equivalent:
t[t[x1, 0], 0]==t[1, 0]
Proof
Note that the input for the rule:
t[p[1, x1_], 0]->t[x1, 0]
contains a subpattern of the form:
p[1, x1_]
which can be unified with the input for the rule:
p[1, t[x1_, 0]]->1
where these rules follow from Critical Pair Lemma 5 and Substitution Lemma 10 respectively.
Substitution Lemma 11
It can be shown that:
t[x1, t[0, 0]]==t[1, 0]
Proof
We start by taking Critical Pair Lemma 17, and apply the substitution:
t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]
which follows from Axiom 9.
Substitution Lemma 12
It can be shown that:
t[x1, 0]==t[1, 0]
Proof
We start by taking Substitution Lemma 11, and apply the substitution:
t[0, 0]->0
which follows from Substitution Lemma 1.
Substitution Lemma 13
It can be shown that:
t[x1, 0]==0
Proof
We start by taking Substitution Lemma 12, and apply the substitution:
t[1, x1_]->x1
which follows from Axiom 3.
Critical Pair Lemma 18
The following expressions are equivalent:
t[x1, t[0, x2]]==t[0, x2]
Proof
Note that the input for the rule:
t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]
contains a subpattern of the form:
t[x1_, x2_]
which can be unified with the input for the rule:
t[x1_, 0]->0
where these rules follow from Axiom 9 and Substitution Lemma 13 respectively.
Critical Pair Lemma 19
The following expressions are equivalent:
p[1, t[0, x1]]==p[n[x1], p[x1, 1]]
Proof
Note that the input for the rule:
p[n[x1_], p[x1_, x2_]]->x2
contains a subpattern of the form:
p[x1_, x2_]
which can be unified with the input for the rule:
p[x1_, p[1, t[0, x1_]]]->p[x1, 1]
where these rules follow from Critical Pair Lemma 13 and Substitution Lemma 7 respectively.
Substitution Lemma 14
It can be shown that:
p[1, t[0, x1]]==1
Proof
We start by taking Critical Pair Lemma 19, and apply the substitution:
p[n[x1_], p[x1_, x2_]]->x2
which follows from Critical Pair Lemma 13.
Critical Pair Lemma 20
The following expressions are equivalent:
t[0, t[0, x1]]==t[0, 1]
Proof
Note that the input for the rule:
t[0, p[1, x1_]]->t[0, x1]
contains a subpattern of the form:
p[1, x1_]
which can be unified with the input for the rule:
p[1, t[0, x1_]]->1
where these rules follow from Critical Pair Lemma 8 and Substitution Lemma 14 respectively.
Substitution Lemma 15
It can be shown that:
t[0, x1]==t[0, 1]
Proof
We start by taking Critical Pair Lemma 20, and apply the substitution:
t[x1_, t[0, x2_]]->t[0, x2]
which follows from Critical Pair Lemma 18.
Substitution Lemma 16
It can be shown that:
t[0, x1]==0
Proof
We start by taking Substitution Lemma 15, and apply the substitution:
t[x1_, 1]->x1
which follows from Axiom 2.
Critical Pair Lemma 21
The following expressions are equivalent:
t[n[1], x1]==p[n[x1], t[0, x1]]
Proof
Note that the input for the rule:
p[n[x1_], t[p[1, x2_], x1_]]->t[x2, x1]
contains a subpattern of the form:
p[1, x2_]
which can be unified with the input for the rule:
p[x1_, n[x1_]]->0
where these rules follow from Critical Pair Lemma 14 and Axiom 6 respectively.
Substitution Lemma 17
It can be shown that:
t[n[1], x1]==p[n[x1], 0]
Proof
We start by taking Critical Pair Lemma 21, and apply the substitution:
t[0, x1_]->0
which follows from Substitution Lemma 16.
Substitution Lemma 18
It can be shown that:
t[n[1], x1]==n[x1]
Proof
We start by taking Substitution Lemma 17, and apply the substitution:
p[x1_, 0]->x1
which follows from Axiom 1.
Critical Pair Lemma 22
The following expressions are equivalent:
t[n[1], t[x1, x2]]==t[n[x1], x2]
Proof
Note that the input for the rule:
t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]
contains a subpattern of the form:
t[x1_, x2_]
which can be unified with the input for the rule:
t[n[1], x1_]->n[x1]
where these rules follow from Axiom 9 and Substitution Lemma 18 respectively.
Substitution Lemma 19
It can be shown that:
n[t[x1, x2]]==t[n[x1], x2]
Proof
We start by taking Critical Pair Lemma 22, and apply the substitution:
t[n[1], x1_]->n[x1]
which follows from Substitution Lemma 18.
Critical Pair Lemma 23
The following expressions are equivalent:
t[x1, n[1]]==p[n[x1], t[x1, 0]]
Proof
Note that the input for the rule:
p[n[x1_], t[x1_, p[1, x2_]]]->t[x1, x2]
contains a subpattern of the form:
p[1, x2_]
which can be unified with the input for the rule:
p[x1_, n[x1_]]->0
where these rules follow from Critical Pair Lemma 15 and Axiom 6 respectively.
Substitution Lemma 20
It can be shown that:
t[x1, n[1]]==p[n[x1], 0]
Proof
We start by taking Critical Pair Lemma 23, and apply the substitution:
t[x1_, 0]->0
which follows from Substitution Lemma 13.
Substitution Lemma 21
It can be shown that:
t[x1, n[1]]==n[x1]
Proof
We start by taking Substitution Lemma 20, and apply the substitution:
p[x1_, 0]->x1
which follows from Axiom 1.
Critical Pair Lemma 24
The following expressions are equivalent:
n[t[x1, x2]]==t[x1, t[x2, n[1]]]
Proof
Note that the input for the rule:
t[x1_, n[1]]->n[x1]
contains a subpattern of the form:
t[x1_, n[1]]
which can be unified with the input for the rule:
t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]
where these rules follow from Substitution Lemma 21 and Axiom 9 respectively.
Substitution Lemma 22
It can be shown that:
t[n[x1], x2]==t[x1, t[x2, n[1]]]
Proof
We start by taking Critical Pair Lemma 24, and apply the substitution:
n[t[x1_, x2_]]->t[n[x1], x2]
which follows from Substitution Lemma 19.
Substitution Lemma 23
It can be shown that:
t[n[x1], x2]==t[x1, n[x2]]
Proof
We start by taking Substitution Lemma 22, and apply the substitution:
t[x1_, n[1]]->n[x1]
which follows from Substitution Lemma 21.
Substitution Lemma 24
It can be shown that:
t[a, n[n[a]]]==t[a, a]
Proof
We start by taking Hypothesis 1, and apply the substitution:
t[n[x1_], x2_]->t[x1, n[x2]]
which follows from Substitution Lemma 23.
Conclusion 1
We obtain the conclusion:
True
Proof
Take Substitution Lemma 24, and apply the substitution:
n[n[x1_]]->x1
which follows from Substitution Lemma 9.