Proof step 1 from proof wand-flatten (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 wand-flatten (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 wand-flatten (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 26:

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

Rule argument

Additional information

Proof step 4 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

  ⊦ ((hP0 -* hP1))(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 6:

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

Rule argument

Additional information

Used terms: h, hP2, hP0 * hP1

Proof step 5 from proof wand-flatten (previous step) (back to tree)

Lemma wand

Proof step 6 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

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

Rule argument

Additional information

Proof step 7 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ ((hP0 -* hP1))(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 9:

  ((hP0 * hP1))(h1), h1 ⊥ h, ¬ hP2(h1 ∪ h), ∀ h1. h1 ⊥ h ∧ hP0(h1) → ((hP1 -* hP2))(h1 ∪ h) ⊦

Rule argument

Additional information

Used terms: h, hP1 -* hP2, hP0

Proof step 8 from proof wand-flatten (previous step) (back to tree)

Lemma wand

Proof step 9 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

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

Rule argument

Additional information

Used terms: h1, hP1, hP0

Proof step 10 from proof wand-flatten (previous step) (back to tree)

Lemma sep

Proof step 11 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  hP0(h0), hP1(h2), h0 ⊥ h2, h ⊥ h0, h ⊥ h2, ¬ hP2((h0 ∪ h2) ∪ h), ∀ h1. h1 ⊥ h ∧ hP0(h1) → ((hP1 -* hP2))(h1 ∪ h) ⊦

Rule argument

Additional information

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

Proof step 12 from proof wand-flatten (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

Used terms: h0

Proof step 13 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s)

Proof step 14 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

  ⊦ ((hP0 -* hP1))(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 16:

  hP0(h0), hP1(h2), h ⊥ h2, h ⊥ h0, h0 ⊥ h2, ¬ hP2((h0 ∪ h2) ∪ h), ∀ h1. h1 ⊥ h0 ∪ h ∧ hP1(h1) → hP2(h1 ∪ h0 ∪ h) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h, hP2, hP1

Proof step 15 from proof wand-flatten (previous step) (back to tree)

Lemma wand

Proof step 16 from proof wand-flatten (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 17:

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

Rule argument

Additional information

Used terms: h2

Proof step 17 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

Proof step 18 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

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

• Proof step 20:

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

Rule argument

Additional information

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

Proof step 19 from proof wand-flatten (previous step) (back to tree)

A spec-lemma

Proof step 20 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s)

Proof step 21 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 23:

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

Rule argument

Additional information

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

Proof step 22 from proof wand-flatten (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s)

Proof step 24 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

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

Rule argument

Additional information

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

Proof step 25 from proof wand-flatten (previous step) (back to tree)

A spec-lemma

Proof step 26 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 27:

  ⊦ ((hP0 -* hP1))(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 28:

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

Rule argument

Additional information

Used terms: h, hP1 -* hP2, hP0

Proof step 27 from proof wand-flatten (previous step) (back to tree)

Lemma wand

Proof step 28 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

Proof step 29 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 30:

  ⊦ ((hP0 -* hP1))(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 31:

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

Rule argument

Additional information

Used terms: h1 ∪ h, hP2, hP1

Proof step 30 from proof wand-flatten (previous step) (back to tree)

Lemma wand

Proof step 31 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

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

Rule argument

Additional information

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

Proof step 32 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 33:

  ⊦ ((hP0 -* hP1))(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 34:

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

Rule argument

Additional information

Used terms: h, hP2, hP0 * hP1

Proof step 33 from proof wand-flatten (previous step) (back to tree)

Lemma wand

Proof step 34 from proof wand-flatten (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 35:

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

Rule argument

Additional information

Used terms: h0 ∪ h1

Proof step 35 from proof wand-flatten (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

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

Rule argument

Additional information

Used simplifier rules: split-union: ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → ((hP0 * hP1))(h0 ∪ h1); sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; sep-com: ⊦ hP0 * hP1 = hP1 * hP0;

Proof step 36 from proof wand-flatten (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 37:

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

Rule argument

Additional information

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

Proof step 37 from proof wand-flatten (previous step) (back to tree)

A spec-lemma