Proof step 1 from proof f-rm (conclusion) (back to tree)

⊦ x.distinct intersect y = (x intersect y).distinct

System applied rule normalize

And obtained the following new goals

Proof step 2:
x.distinct intersect y ≠ (x intersect y).distinct ⊦

Rule argument

Additional information

Proof step 2 from proof f-rm (previous step) (back to tree)

x.distinct intersect y ≠ (x intersect y).distinct ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 3:
(a + x0).distinct intersect y ≠ ((a + x0) intersect y).distinct, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

indvar: x

Additional information

Used simplifier rules: inters-e: ⊦ [] intersect y = []; List('a) freely generated by [], +;; ⊦ [].distinct = []; (s)

Proof step 3 from proof f-rm (previous step) (back to tree)

(a + x0).distinct intersect y ≠ ((a + x0) intersect y).distinct, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(a ' + x0).distinct intersect y ≠ ((a ' + x0) intersect y).distinct, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)

Proof step 4 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ ((a ' + x0) intersect y).distinct, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
(a ' + x0).distinct intersect y ≠ ((a ' + x0) intersect y).distinct, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦
Proof step 17:
(a ' + x0).distinct intersect y ≠ ((a ' + x0) intersect y).distinct, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y

Proof step 5 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ ((a ' + x0) intersect y).distinct, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(a ' + x0).distinct intersect y ≠ (a ' + (x0 intersect y)).distinct, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: inters-y: ⊦ a ∈ y → (a ' + x) intersect y = a ' + (x intersect y); a: ⊦ (x + y) + z = x + y + z; from specification polylist

Proof step 6 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ (a ' + (x0 intersect y)).distinct, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ x → (a ' + x).distinct = x.distinct
Proof step 8:
(a ' + x0).distinct intersect y ≠ (a ' + (x0 intersect y)).distinct, a ∈ y, ¬ a ∈ x0 intersect y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦
Proof step 13:
(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, a ∈ y, a ∈ x0 intersect y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

spec: polylist-dup, name: rmdup-y, seq: ⊦ a ∈ x → (a ' + x).distinct = x.distinct, subst:[x, a] → [x0 intersect y, a], <all paths>

Additional information

Used terms: x0 intersect y, a

Proof step 7 from proof f-rm (previous step) (back to tree)

⊦ a ∈ x → (a ' + x).distinct = x.distinct

A spec-lemma

Proof step 8 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ (a ' + (x0 intersect y)).distinct, a ∈ y, ¬ a ∈ x0 intersect y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
(a ' + x0).distinct intersect y ≠ a ' + (x0 intersect y).distinct, a ∈ y, ¬ a ∈ x0 intersect y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ a + x = a ' + x; (s) ⊦ ¬ a ∈ x → (a ' + x).distinct = a + x.distinct; (s)

Proof step 9 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ a ' + (x0 intersect y).distinct, a ∈ y, ¬ a ∈ x0 intersect y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ∈ x intersect y ↔ a ∈ x ∧ a ∈ y
Proof step 11:
(a ' + x0).distinct intersect y ≠ a ' + (x0 intersect y).distinct, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct, ¬ a ∈ x0 ∨ ¬ a ∈ y ⊦

Rule argument

spec: <toplevel>, name: inters-in, seq: ⊦ a ∈ x intersect y ↔ a ∈ x ∧ a ∈ y, subst:[y, x, a] → [y, x0, a], <all paths>

Additional information

Used terms: y, x0, a

Proof step 10 from proof f-rm (previous step) (back to tree)

⊦ a ∈ x intersect y ↔ a ∈ x ∧ a ∈ y

Lemma inters-in

Proof step 11 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ a ' + (x0 intersect y).distinct, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct, ¬ a ∈ x0 ∨ ¬ a ∈ y ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
x0.distinct intersect y ≠ (x0 intersect y).distinct, a ∈ y, ¬ a ∈ x0, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: inters-y: ⊦ a ∈ y → (a ' + x) intersect y = a ' + (x intersect y); a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ x + y = x + z ↔ y = z; (s) ⊦ a + x = a ' + x; (s) ⊦ ¬ a ∈ x → (a ' + x).distinct = a + x.distinct; (s)

Proof step 12 from proof f-rm (previous step) (back to tree)

x0.distinct intersect y ≠ (x0 intersect y).distinct, a ∈ y, ¬ a ∈ x0, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: y

Additional information

Used terms: y

Proof step 13 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, a ∈ y, a ∈ x0 intersect y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ a ∈ x intersect y ↔ a ∈ x ∧ a ∈ y
Proof step 15:
(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, a ∈ x0, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

spec: <toplevel>, name: inters-in, seq: ⊦ a ∈ x intersect y ↔ a ∈ x ∧ a ∈ y, subst:[y, x, a] → [y, x0, a], <all paths>

Additional information

Used terms: y, x0, a

Proof step 14 from proof f-rm (previous step) (back to tree)

⊦ a ∈ x intersect y ↔ a ∈ x ∧ a ∈ y

Lemma inters-in

Proof step 15 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, a ∈ x0, a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
x0.distinct intersect y ≠ (x0 intersect y).distinct, a ∈ y, a ∈ x0, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ a ∈ x → (a ' + x).distinct = x.distinct; (s)

Proof step 16 from proof f-rm (previous step) (back to tree)

x0.distinct intersect y ≠ (x0 intersect y).distinct, a ∈ y, a ∈ x0, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: y

Additional information

Used terms: y

Proof step 17 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ ((a ' + x0) intersect y).distinct, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: inters-n: ⊦ ¬ a ∈ y → (a ' + x) intersect y = x intersect y; a: ⊦ (x + y) + z = x + y + z; from specification polylist

Proof step 18 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ a ∈ x → (a ' + x).distinct = x.distinct
Proof step 20:
(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, ¬ a ∈ x0, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦
Proof step 22:
x0.distinct intersect y ≠ (x0 intersect y).distinct, a ∈ x0, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

spec: polylist-dup, name: rmdup-y, seq: ⊦ a ∈ x → (a ' + x).distinct = x.distinct, subst:[x, a] → [x0, a], <all paths>

Additional information

Used terms: x0, a

Proof step 19 from proof f-rm (previous step) (back to tree)

⊦ a ∈ x → (a ' + x).distinct = x.distinct

A spec-lemma

Proof step 20 from proof f-rm (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ (x0 intersect y).distinct, ¬ a ∈ x0, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
x0.distinct intersect y ≠ (x0 intersect y).distinct, ¬ a ∈ x0, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Rule argument

Additional information

Used simplifier rules: inters-n: ⊦ ¬ a ∈ y → (a ' + x) intersect y = x intersect y; a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ a + x = a ' + x; (s) ⊦ ¬ a ∈ x → (a ' + x).distinct = a + x.distinct; (s)

Proof step 21 from proof f-rm (previous step) (back to tree)

x0.distinct intersect y ≠ (x0 intersect y).distinct, ¬ a ∈ x0, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: y

Additional information

Used terms: y

Proof step 22 from proof f-rm (previous step) (back to tree)

x0.distinct intersect y ≠ (x0 intersect y).distinct, a ∈ x0, ¬ a ∈ y, ∀ y. x0.distinct intersect y = (x0 intersect y).distinct ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: y

Additional information

Used terms: y

Proof Tree for f-rm (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.