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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  precise(hP0), precise(hP1), ¬ precise(hP0 * hP1) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  precise(hP1), precise(hP0), ¬ precise(hP0 * hP1) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ precise(hP) ↔ (∀ h, h0, h1, h2. hP(h) ∧ hP(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0)

• Proof step 5:

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

Rule argument

Additional information

Used terms: hP0 * hP1

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

Lemma precise

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, ((hP0 * hP1))(h), ((hP0 * hP1))(h0), precise(hP1), precise(hP0), h ⊥ h1, h0 ⊥ h2 ⊦

Rule argument

Additional information

Proof step 6 from proof precise-sep (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:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, ((hP0 * hP1))(h), precise(hP0), precise(hP1), h0 ⊥ h2, h ⊥ h1, ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ hP0(h1) ∧ hP1(h2) ⊦

Rule argument

Additional information

Used terms: h0, hP1, hP0

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  h ∪ h1 = (h3 ∪ h4) ∪ h2, h ≠ h3 ∪ h4, ((hP0 * hP1))(h), hP0(h3), hP1(h4), precise(hP1), precise(hP0), h3 ⊥ h4, h ⊥ h1, h2 ⊥ h3, h2 ⊥ h4 ⊦

Rule argument

Additional information

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

Proof step 9 from proof precise-sep (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:

  h ∪ h1 = (h3 ∪ h4) ∪ h2, h ≠ h3 ∪ h4, hP0(h3), hP1(h4), precise(hP0), precise(hP1), h2 ⊥ h4, h2 ⊥ h3, h ⊥ h1, h3 ⊥ h4, ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1) ⊦

Rule argument

Additional information

Used terms: h, hP1, hP0

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

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

Rule argument

Additional information

Used simplifier rules: precise-same: precise(hP) ⊦ hP(h0) ∧ hP(h1) ∧ h0 ⊥ h ∧ h1 ⊥ h2 → (h0 ∪ h = h1 ∪ h2 ↔ h0 = h1 ∧ h = h2); ⊦ h1 ⊥ h2 → (h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2); (ws)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ h0 ⊥ h1 ↔ h1 ⊥ h0

• Proof step 14:

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

Rule argument

Additional information

Used terms: h5, h0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ h0 ⊥ h1 ↔ h1 ⊥ h0

• Proof step 17:

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

Rule argument

Additional information

Used terms: h3, h2

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ precise(hP) ↔ (∀ h, h0, h1, h2. hP(h) ∧ hP(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0)

• Proof step 20:

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

Rule argument

Additional information

Used terms: hP0

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

Lemma precise

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

  ⊦ precise(hP) ↔ (∀ h, h0, h1, h2. hP(h) ∧ hP(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0)

• Proof step 22:

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

Rule argument

Additional information

Used terms: hP1

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

Lemma precise

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

User applied rule all left

And obtained the following new goals

• Proof step 23:

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

Rule argument

Additional information

Used terms: h0, h3, h5 ∪ h1, h4 ∪ h2

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

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

Rule argument

Additional information

Proof step 24 from proof precise-sep (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

• Proof step 26:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h1, h5, h0

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

A spec-lemma

Proof step 26 from proof precise-sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

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

Rule argument

Additional information

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

Proof step 27 from proof precise-sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 28:

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

• Proof step 29:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h2, h4, h3

Proof step 28 from proof precise-sep (previous step) (back to tree)

A spec-lemma

Proof step 29 from proof precise-sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

Rule argument

Additional information

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

Proof step 30 from proof precise-sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 31:

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

• Proof step 32:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h1, h5, h0

Proof step 31 from proof precise-sep (previous step) (back to tree)

A spec-lemma

Proof step 32 from proof precise-sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

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

Rule argument

Additional information

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

Proof step 33 from proof precise-sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

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

• Proof step 35:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h2, h4, h3

Proof step 34 from proof precise-sep (previous step) (back to tree)

A spec-lemma

Proof step 35 from proof precise-sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

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

Rule argument

Additional information

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

Proof step 36 from proof precise-sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 37:

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

• Proof step 38:

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

• Proof step 43:

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

Rule argument

Additional information

Used terms: h4 ∪ h2, h5 ∪ h1, h3

Proof step 37 from proof precise-sep (previous step) (back to tree)

A spec-lemma

Proof step 38 from proof precise-sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 39:

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

• Proof step 40:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h1, h5, h3

Proof step 39 from proof precise-sep (previous step) (back to tree)

A spec-lemma

Proof step 40 from proof precise-sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

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

Rule argument

Additional information

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

Proof step 41 from proof precise-sep (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h2, h4, h3

Proof step 42 from proof precise-sep (previous step) (back to tree)

A spec-lemma

Proof step 43 from proof precise-sep (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

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

Rule argument

Additional information

Proof step 44 from proof precise-sep (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; (s) Used terms: h5, h4, h1, h2