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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  hP0 ⊗ hP1 ≠ hP1 ⊗ hP0 ⊦

Rule argument

Additional information

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

User applied rule expand left

And obtained the following new goals

• Proof step 3:

  ¬((hP0 ⊗ hP1)(h) ↔ (hP1 ⊗ hP0)(h)) ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

  (hP0 ⊗ hP1)(h), ¬ (hP1 ⊗ hP0)(h) ⊦

• Proof step 10:

  (hP1 ⊗ hP0)(h), ¬ (hP0 ⊗ hP1)(h) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

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

• Proof step 6:

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

Rule argument

Additional information

Used terms: h, hP1, hP0

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

Lemma sharing

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (hP1 ⊗ hP0)(h0 ∪ h1 ∪ h2) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

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

• Proof step 9:

  h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ∀ h3, h4, h5. ¬(h3 ⊥ h5 ∧ h3 ⊥ h4 ∧ h4 ⊥ h5 ∧ h0 ∪ h1 ∪ h2 = h3 ∪ h4 ∪ h5 ∧ hP1(h3 ∪ h5) ∧ hP0(h4 ∪ h5)) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h1 ∪ h2, hP0, hP1

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

Lemma sharing

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) h0 ⊥ h1, h1 ⊥ h, h ⊥ h0 ⊦ h ∪ h0 ∪ h1 = h0 ∪ h ∪ h1 ↔ true; (ss) Used terms: h1, h0, h2

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

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

• Proof step 12:

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

Rule argument

Additional information

Used terms: h, hP1, hP0

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

Lemma sharing

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  (hP1 ⊗ hP0)(h), ∀ h0, h1, h2. ¬ h0 ⊥ h2 ∨ ¬ h0 ⊥ h1 ∨ ¬ h1 ⊥ h2 ∨ h ≠ h0 ∪ h1 ∪ h2 ∨ ¬ hP0(h0 ∪ h2) ∨ ¬ hP1(h1 ∪ h2) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

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

• Proof step 15:

  ∀ h0, h1, h2. ¬ h0 ⊥ h2 ∨ ¬ h0 ⊥ h1 ∨ ¬ h1 ⊥ h2 ∨ h ≠ h0 ∪ h1 ∪ h2 ∨ ¬ hP0(h0 ∪ h2) ∨ ¬ hP1(h1 ∪ h2), ∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP1(h0 ∪ h2) ∧ hP0(h1 ∪ h2) ⊦

Rule argument

Additional information

Used terms: h, hP0, hP1

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

Lemma sharing

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP0(h1 ∪ h2), hP1(h0 ∪ h2), ∀ h3, h4, h5. ¬ h3 ⊥ h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h4 ⊥ h5 ∨ h0 ∪ h1 ∪ h2 ≠ h3 ∪ h4 ∪ h5 ∨ ¬ hP0(h3 ∪ h5) ∨ ¬ hP1(h4 ∪ h5) ⊦

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) h0 ⊥ h1, h1 ⊥ h, h ⊥ h0 ⊦ h ∪ h0 ∪ h1 = h0 ∪ h ∪ h1 ↔ true; (ss) Used terms: h1, h0, h2