Proof step 1 from proof sharing-emp (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  hP ⊗ emp ≠ hP ⊦

Rule argument

Additional information

Proof step 2 from proof sharing-emp (previous step) (back to tree)

User applied rule expand left

And obtained the following new goals

• Proof step 3:

  ¬(((hP ⊗ emp))(h) ↔ hP(h)) ⊦

Rule argument

Additional information

Proof step 3 from proof sharing-emp (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ ((hP0 ⊗ hP1))(h) ↔ (∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP0(h0 ∪ h2) ∧ hP1(h1 ∪ h2))

• Proof step 5:

  ⊦ (∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP(h0 ∪ h2) ∧ emp(h1 ∪ h2)) ↔ hP(h)

Rule argument

Additional information

Used terms: h, emp, hP

Proof step 4 from proof sharing-emp (previous step) (back to tree)

Lemma sharing

Proof step 5 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ⊦ (∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP(h0 ∪ h2) ∧ h1 ∪ h2 = ∅) ↔ hP(h)

Rule argument

Additional information

Used simplifier rules: emp: ⊦ emp(h) ↔ h = ∅;

Proof step 6 from proof sharing-emp (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

  ¬ hP(h), ∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP(h0 ∪ h2) ∧ h1 ∪ h2 = ∅ ⊦

• Proof step 23:

  hP(h) ⊦ ∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP(h0 ∪ h2) ∧ h1 ∪ h2 = ∅

Rule argument

Additional information

Proof step 7 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  h1 ∪ h2 = ∅, hP(h0 ∪ h2), h1 ⊥ h2, h0 ⊥ h1, h0 ⊥ h2, ¬ hP(h0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ∪ ∅ = h0; (s)

Proof step 8 from proof sharing-emp (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 9:

  h1 ∪ h2 = ∅, hP(h0 ∪ h2), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ¬ hP(h0) ⊦ h1 = ∅ ∧ h2 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ∅ ∪ h = h; (s) ⊦ ∅ ⊥ h; (s) ⊦ h ⊥ ∅; (s) ⊦ h0 ∪ ∅ = h0; (s) Used formulas: h1 = ∅ ∧ h2 = ∅

Proof step 9 from proof sharing-emp (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0[r] = h1[r]))

• Proof step 11:

  hP(h0 ∪ h2), h1 ⊥ h2, h0 ⊥ h1, h0 ⊥ h2, ¬ hP(h0), ∀ r. (r ∈ h1 ∪ h2 ↔ r ∈ ∅) ∧ (r ∈ h1 ∪ h2 → (h1 ∪ h2)[r] = ∅[r]) ⊦ h1 = ∅ ∧ h2 = ∅

Rule argument

Additional information

Used terms: ∅, h1 ∪ h2

Proof step 10 from proof sharing-emp (previous step) (back to tree)

A spec-lemma

Proof step 11 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  hP(h0 ∪ h2), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦ h1 = ∅ ∧ h2 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) ⊦ ¬ r ∈ ∅; (s)

Proof step 12 from proof sharing-emp (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0[r] = h1[r]))

• Proof step 14:

  hP(h0 ∪ h2), h1 ⊥ h2, h0 ⊥ h1, h0 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦ (∀ r. (r ∈ h1 ↔ r ∈ ∅) ∧ (r ∈ h1 → h1[r] = ∅[r])) ∧ h2 = ∅

Rule argument

Additional information

Used terms: ∅, h1

Proof step 13 from proof sharing-emp (previous step) (back to tree)

A spec-lemma

Proof step 14 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  hP(h0 ∪ h2), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦ ¬ r ∈ h1 ∧ h2 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ r ∈ ∅; (s)

Proof step 15 from proof sharing-emp (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0[r] = h1[r]))

• Proof step 17:

  hP(h0 ∪ h2), h1 ⊥ h2, h0 ⊥ h1, h0 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦ ¬ r ∈ h1 ∧ (∀ r. (r ∈ h2 ↔ r ∈ ∅) ∧ (r ∈ h2 → h2[r] = ∅[r]))

Rule argument

Additional information

Used terms: ∅, h2

Proof step 16 from proof sharing-emp (previous step) (back to tree)

A spec-lemma

Proof step 17 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  hP(h0 ∪ h2), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦ ¬ r ∈ h1 ∧ ¬ r0 ∈ h2

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ r ∈ ∅; (s)

Proof step 18 from proof sharing-emp (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 19:

  hP(h0 ∪ h2), r ∈ h1, h1 ⊥ h2, h0 ⊥ h1, h0 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦

• Proof step 21:

  hP(h0 ∪ h2), r0 ∈ h2, h1 ⊥ h2, h0 ⊥ h1, h0 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦

Rule argument

Additional information

Proof step 19 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  hP(h0 ∪ h2), r ∈ h1, h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦

Rule argument

Additional information

Proof step 20 from proof sharing-emp (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r

Proof step 21 from proof sharing-emp (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  hP(h0 ∪ h2), r0 ∈ h2, h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ¬ hP(h0), ∀ r. ¬ r ∈ h1 ∧ ¬ r ∈ h2 ⊦

Rule argument

Additional information

Proof step 22 from proof sharing-emp (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r0

Proof step 23 from proof sharing-emp (previous step) (back to tree)

User applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ h ⊥ ∅; (s) ⊦ ∅ ⊥ h; (s) ⊦ h0 ∪ ∅ = h0; (s) ⊦ ∅ ∪ h = h; (s) Used terms: h, ∅, ∅