Proof step 1 from proof le (conclusion) (back to tree)

⊦ #(x.distinct intersect y) ≤ # y

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬ #(x.distinct intersect y) ≤ # y ⊦

Rule argument

Additional information

Proof step 2 from proof le (previous step) (back to tree)

¬ #(x.distinct intersect y) ≤ # y ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x.distinct intersect y ≠ [], # y < #(x.distinct intersect y) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 3 from proof le (previous step) (back to tree)

x.distinct intersect y ≠ [], # y < #(x.distinct intersect y) ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
(a + x0).distinct intersect y ≠ [], # y < #((a + x0).distinct intersect y), ∀ y. ¬(x0.distinct intersect y ≠ [] ∧ # y < #(x0.distinct intersect y)) ⊦

Rule argument

indvar: x

Additional information

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

Proof step 4 from proof le (previous step) (back to tree)

(a + x0).distinct intersect y ≠ [], # y < #((a + x0).distinct intersect y), ∀ y. ¬(x0.distinct intersect y ≠ [] ∧ # y < #(x0.distinct intersect y)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(a ' + x0).distinct intersect y ≠ [], # y < #((a ' + x0).distinct intersect y), ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ a + x = a ' + x; (s)

Proof step 5 from proof le (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ [], # y < #((a ' + x0).distinct intersect y), ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ x → (a ' + x).distinct = x.distinct
Proof step 7:
(a ' + x0).distinct intersect y ≠ [], # y < #((a ' + x0).distinct intersect y), ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦
Proof step 14:
x0.distinct intersect y ≠ [], # y < #(x0.distinct intersect y), a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

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 6 from proof le (previous step) (back to tree)

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

A spec-lemma

Proof step 7 from proof le (previous step) (back to tree)

(a ' + x0).distinct intersect y ≠ [], # y < #((a ' + x0).distinct intersect y), ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
(a ' + x0.distinct) intersect y ≠ [], # y < #((a ' + x0.distinct) intersect y), ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic 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 8 from proof le (previous step) (back to tree)

(a ' + x0.distinct) intersect y ≠ [], # y < #((a ' + x0.distinct) intersect y), ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 9:
(a ' + x0.distinct) intersect y ≠ [], # y < #((a ' + x0.distinct) intersect y), a ∈ y, ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦
Proof step 12:
(a ' + x0.distinct) intersect y ≠ [], # y < #((a ' + x0.distinct) intersect y), ¬ a ∈ y, ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y

Proof step 9 from proof le (previous step) (back to tree)

(a ' + x0.distinct) intersect y ≠ [], # y < #((a ' + x0.distinct) intersect y), a ∈ y, ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a ∈ y, # y ≤ #(x0.distinct intersect y), ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Rule argument

Additional information

Used simplifier rules: inters-y: ⊦ a ∈ y → (a ' + x) intersect y = a ' + (x intersect y); ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ # a ' = 0 +1; (s) ⊦ #(x + y) = # x + # y; (s) ⊦ a ' + x ≠ []; (s)

Proof step 10 from proof le (previous step) (back to tree)

a ∈ y, # y ≤ #(x0.distinct intersect y), ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

User applied rule all left

And obtained the following new goals

Proof step 11:
a ∈ y, # y ≤ #(x0.distinct intersect y), ¬ a ∈ x0, x0.distinct intersect (y -1l a) = [] ∨ ¬ #(y -1l a) < #(x0.distinct intersect (y -1l a)) ⊦

Rule argument

pos: left 4 terms: y -1l a abort: F discard: T computed: F

Additional information

Used terms: y -1l a

Proof step 11 from proof le (previous step) (back to tree)

a ∈ y, # y ≤ #(x0.distinct intersect y), ¬ a ∈ x0, x0.distinct intersect (y -1l a) = [] ∨ ¬ #(y -1l a) < #(x0.distinct intersect (y -1l a)) ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del1-01: ⊦ ¬ a ∈ x → x.distinct intersect (y -1l a) = x.distinct intersect y; intersect-empty-01: ⊦ x intersect [] = []; ⊦ ¬ a ∈ []; (s) ⊦ # x = 0 ↔ x = []; (s) ⊦ m ≤ 0 ↔ m = 0; (s) ⊦ # [] = 0; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ n -1 = n - 1; (s) ⊦ a ∈ x → #(x -1l a) = (# x)-1; (s)

Proof step 12 from proof le (previous step) (back to tree)

(a ' + x0.distinct) intersect y ≠ [], # y < #((a ' + x0.distinct) intersect y), ¬ a ∈ y, ¬ a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
x0.distinct intersect y ≠ [], # y < #(x0.distinct intersect y), ¬ a ∈ x0, ¬ a ∈ y, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

Rule argument

Additional information

Used simplifier rules: inters-n: ⊦ ¬ a ∈ y → (a ' + x) intersect y = x intersect y; ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic a: ⊦ (x + y) + z = x + y + z; from specification polylist

Proof step 13 from proof le (previous step) (back to tree)

x0.distinct intersect y ≠ [], # y < #(x0.distinct intersect y), ¬ a ∈ x0, ¬ a ∈ y, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: y abort: F discard: T computed: F

Additional information

Used terms: y

Proof step 14 from proof le (previous step) (back to tree)

x0.distinct intersect y ≠ [], # y < #(x0.distinct intersect y), a ∈ x0, ∀ y0. x0.distinct intersect y0 = [] ∨ ¬ # y0 < #(x0.distinct intersect y0) ⊦

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 le (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.