Proof step 1 from proof sep-ass (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  (hP0 * hP1) * hP2 ≠ hP0 * hP1 * hP2 ⊦

Rule argument

Additional information

Proof step 2 from proof sep-ass (previous step) (back to tree)

User applied rule expand left

And obtained the following new goals

• Proof step 3:

  ¬(((hP0 * hP1) * hP2)(h) ↔ (hP0 * hP1 * hP2)(h)) ⊦

Rule argument

Additional information

Proof step 3 from proof sep-ass (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

  ((hP0 * hP1) * hP2)(h), ¬ (hP0 * hP1 * hP2)(h) ⊦

• Proof step 15:

  (hP0 * hP1 * hP2)(h), ¬ ((hP0 * hP1) * hP2)(h) ⊦

Rule argument

Additional information

Proof step 4 from proof sep-ass (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. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 6:

  ¬ (hP0 * hP1 * hP2)(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ (hP0 * hP1)(h0) ∧ hP2(h1) ⊦

Rule argument

Additional information

Used terms: h, hP2, hP0 * hP1

Proof step 5 from proof sep-ass (previous step) (back to tree)

Lemma sep

Proof step 6 from proof sep-ass (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  h0 ⊥ h1, (hP0 * hP1)(h0), hP2(h1), ¬ (hP0 * hP1 * hP2)(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 7 from proof sep-ass (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. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 9:

  h0 ⊥ h1, hP2(h1), ¬ (hP0 * hP1 * hP2)(h0 ∪ h1), ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ hP0(h1) ∧ hP1(h2) ⊦

Rule argument

Additional information

Used terms: h0, hP1, hP0

Proof step 8 from proof sep-ass (previous step) (back to tree)

Lemma sep

Proof step 9 from proof sep-ass (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP0(h), hP1(h2), hP2(h1), ¬ (hP0 * hP1 * hP2)((h ∪ h2) ∪ h1) ⊦

Rule argument

Additional information

Used simplifier rules: h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s)

Proof step 10 from proof sep-ass (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ h0 ⊥ h1 ∧ h1 ⊥ h ∧ h ⊥ h0 → (h0 ∪ h1) ∪ h = h0 ∪ h1 ∪ h

• Proof step 12:

  h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, h2 ⊥ h1, hP0(h), hP1(h2), hP2(h1), ¬ (hP0 * hP1 * hP2)(h ∪ h2 ∪ h1) ⊦

Rule argument

Additional information

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

Proof step 11 from proof sep-ass (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof sep-ass (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, h2 ⊥ h1, hP0(h), hP1(h2), hP2(h1), ¬ (hP0 * hP1 * hP2)(h ∪ h2 ∪ h1) ⊦

Rule argument

Additional information

Proof step 13 from proof sep-ass (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1)

Rule argument

Additional information

Used simplifier rules: split-union: ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1); h1 ⊥ h ⊦ h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h; (ws) ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h2 ∪ h1, h, hP1 * hP2, hP0

Proof step 14 from proof sep-ass (previous step) (back to tree)

Lemma split-union

Proof step 15 from proof sep-ass (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  ¬ ((hP0 * hP1) * hP2)(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ (hP1 * hP2)(h1) ⊦

Rule argument

Additional information

Used terms: h, hP1 * hP2, hP0

Proof step 16 from proof sep-ass (previous step) (back to tree)

Lemma sep

Proof step 17 from proof sep-ass (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  h0 ⊥ h1, hP0(h0), (hP1 * hP2)(h1), ¬ ((hP0 * hP1) * hP2)(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 18 from proof sep-ass (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

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

• Proof step 20:

  h0 ⊥ h1, hP0(h0), ¬ ((hP0 * hP1) * hP2)(h0 ∪ h1), ∃ h0, h2. h0 ⊥ h2 ∧ h1 = h0 ∪ h2 ∧ hP1(h0) ∧ hP2(h2) ⊦

Rule argument

Additional information

Used terms: h1, hP2, hP1

Proof step 19 from proof sep-ass (previous step) (back to tree)

Lemma sep

Proof step 20 from proof sep-ass (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, hP0(h0), hP1(h), hP2(h2), ¬ ((hP0 * hP1) * hP2)(h0 ∪ h ∪ h2) ⊦

Rule argument

Additional information

Used simplifier rules: h1 ⊥ h ⊦ h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h; (ws)

Proof step 21 from proof sep-ass (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ h0 ⊥ h1 ∧ h1 ⊥ h ∧ h ⊥ h0 → (h0 ∪ h1) ∪ h = h0 ∪ h1 ∪ h

• Proof step 23:

  h ⊥ h2, h0 ⊥ h2, h0 ⊥ h, h2 ⊥ h0, hP0(h0), hP1(h), hP2(h2), ¬ ((hP0 * hP1) * hP2)((h0 ∪ h) ∪ h2) ⊦

Rule argument

Additional information

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

Proof step 22 from proof sep-ass (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof sep-ass (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, h2 ⊥ h0, hP0(h0), hP1(h), hP2(h2), ¬ ((hP0 * hP1) * hP2)((h0 ∪ h) ∪ h2) ⊦

Rule argument

Additional information

Proof step 24 from proof sep-ass (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1)

Rule argument

Additional information

Used simplifier rules: split-union: ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1); h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s) ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h2, h0 ∪ h, hP2, hP0 * hP1

Proof step 25 from proof sep-ass (previous step) (back to tree)

Lemma split-union