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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

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

• Proof step 5:

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

Rule argument

Additional information

Used terms: h, hP1, hP2

Proof step 4 from proof sep-partial-mp (previous step) (back to tree)

Lemma wand

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

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

• Proof step 8:

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

Rule argument

Additional information

Used terms: h, hP0 * hP2 -* hP1, hP0

Proof step 7 from proof sep-partial-mp (previous step) (back to tree)

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  ((hP0 * hP2 -* hP1))(h2), hP0(h0), hP2(h1), h0 ⊥ h2, h1 ⊥ h0, h1 ⊥ h2, ¬ hP1(h1 ∪ h0 ∪ h2) ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

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

Rule argument

Additional information

Used terms: h2, hP1, hP0 * hP2

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

Lemma wand

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

User applied rule all left

And obtained the following new goals

• Proof step 12:

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

Rule argument

Additional information

Used terms: h0 ∪ h1

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ h1 ⊥ h2 → (h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2)

• Proof step 14:

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

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  hP0(h0), hP2(h1), h0 ⊥ h2, h1 ⊥ h0, h1 ⊥ h2, ¬ hP1(h1 ∪ h0 ∪ h2), ((hP0 * hP2))(h0 ∪ h1) → hP1((h0 ∪ h1) ∪ h2) ⊦

Rule argument

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

• Proof step 16:

  hP0(h0), hP2(h1), h1 ⊥ h2, h1 ⊥ h0, h0 ⊥ h2, ¬ ((hP0 * hP2))(h0 ∪ h1), ¬ hP1(h1 ∪ h0 ∪ h2) ⊦

• Proof step 18:

  hP0(h0), hP1((h0 ∪ h1) ∪ h2), hP2(h1), h1 ⊥ h2, h1 ⊥ h0, h0 ⊥ h2, ¬ hP1(h1 ∪ h0 ∪ h2) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h1, h0, hP2, hP0

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

Lemma split-union

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 20:

  hP0(h0), hP1((h1 ∪ h0) ∪ h2), hP2(h1), h0 ⊥ h2, h1 ⊥ h2, h1 ⊥ h0, ¬ hP1(h1 ∪ h0 ∪ h2) ⊦

Rule argument

Additional information

Used terms: h1, h0

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

A spec-lemma

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

  ⊦ 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: h2, h0, h1

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

A spec-lemma