Proof step 1 from proof intersect-all (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  x intersect y ≠ [], ∀ a. a ∈ x → ¬ a ∈ y ⊦

Rule argument

Additional information

Proof step 2 from proof intersect-all (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  (a + x0) intersect y ≠ [], ∀ a0. a0 ∈ a + x0 → ¬ a0 ∈ y, ∀ y. ¬(x0 intersect y ≠ [] ∧ (∀ a. a ∈ x0 → ¬ a ∈ y)) ⊦

Rule argument

Additional information

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

Proof step 3 from proof intersect-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  (a ' + x0) intersect y ≠ [], ∀ y0. x0 intersect y0 = [] ∨ ¬(∀ a. a ∈ x0 → ¬ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 → ¬ a0 ∈ y ⊦

Rule argument

Additional information

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

Proof step 4 from proof intersect-all (previous step) (back to tree)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

• Proof step 5:

  (a ' + x0) intersect y ≠ [], a ∈ y, ∀ a0. a0 ∈ a ' + x0 → ¬ a0 ∈ y, ∀ y0. x0 intersect y0 = [] ∨ ¬(∀ a. a ∈ x0 → ¬ a ∈ y0) ⊦

• Proof step 7:

  (a ' + x0) intersect y ≠ [], ¬ a ∈ y, ∀ a0. a0 ∈ a ' + x0 → ¬ a0 ∈ y, ∀ y0. x0 intersect y0 = [] ∨ ¬(∀ a. a ∈ x0 → ¬ a ∈ y0) ⊦

Rule argument

Additional information

Used formulas: a ∈ y

Proof step 5 from proof intersect-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  a ∈ y, ∀ y0. x0 intersect y0 = [] ∨ ¬(∀ a. a ∈ x0 → ¬ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 → ¬ a0 ∈ y ⊦

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 ⊦ a ' + x ≠ []; (s)

Proof step 6 from proof intersect-all (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s) ⊦ a ∈ a0 ' ↔ a = a0; (s) a: ⊦ (x + y) + z = x + y + z; from specification polylist Used terms: a

Proof step 7 from proof intersect-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  x0 intersect y ≠ [], ¬ a ∈ y, ∀ y0. x0 intersect y0 = [] ∨ ¬(∀ a. a ∈ x0 → ¬ a ∈ y0), ∀ a0. a0 ∈ a ' + x0 → ¬ a0 ∈ y ⊦

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 8 from proof intersect-all (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 9:

  x0 intersect y ≠ [], ¬ a ∈ y, ∀ a0. a0 ∈ a ' + x0 → ¬ a0 ∈ y, x0 intersect y = [] ∨ ¬(∀ a. a ∈ x0 → ¬ a ∈ y) ⊦

Rule argument

Additional information

Used terms: y

Proof step 9 from proof intersect-all (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  a0 ∈ y, a0 ∈ x0, ¬ a ∈ y, ∀ a1. a1 ∈ x0 → ¬ a1 ∈ y ⊦

Rule argument

Additional information

Used simplifier rules: intersect-notempty: ⊦ a ∈ x ∧ a ∈ y → x intersect y ≠ []; a: ⊦ (x + y) + z = x + y + z; from specification polylist ⊦ a ∈ a0 ' + x ↔ a = a0 ∨ a ∈ x; (s)

Proof step 10 from proof intersect-all (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: a0