Proof step 1 from proof ex-shift-o (conclusion) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 2:

  ((!(me, hO, _∃ ohP) * hP))(h) ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h)

• Proof step 16:

  ¬ ((!(me, hO, _∃ ohP) * hP))(h), ∃ o. ((!(me, hO, ohP(o)) * hP))(h) ⊦

Rule argument

Additional information

Proof step 2 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

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

• Proof step 4:

  ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ !(me, hO, _∃ ohP)(h0) ∧ hP(h1) ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h)

Rule argument

Additional information

Used terms: h, hP, !(me, hO, _∃ ohP)

Proof step 3 from proof ex-shift-o (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  !(me, hO, _∃ ohP)(h0), hP(h1), h0 ⊥ h1 ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Proof step 5 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

• Proof step 7:

  (_∃ ohP)(h0), hP(h1), h0 ⊥ h1, ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Used terms: h0, _∃ ohP, hO, me

Proof step 6 from proof ex-shift-o (previous step) (back to tree)

Lemma owned

Proof step 7 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ oP)(h) ↔ (∃ o. oP(o)(h)); (s)

Proof step 8 from proof ex-shift-o (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 9:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Used terms: o

Proof step 9 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1), !(me, hO, ohP(o))(h0) ∧ hP(h1) ∧ h0 ⊥ h1

Rule argument

Additional information

Used terms: h1, h0, hP, !(me, hO, ohP(o))

Proof step 10 from proof ex-shift-o (previous step) (back to tree)

A spec-lemma

Proof step 11 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ !(me, hO, ohP(o))(h0), ¬ ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Proof step 12 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

• Proof step 14:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me, ¬ ohP(o)(h0) ∨ (∃ r. ¬(r ∈ h0 → hO(r) = me)) ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Used terms: h0, ohP(o), hO, me

Proof step 13 from proof ex-shift-o (previous step) (back to tree)

Lemma owned

Proof step 14 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  hO(r) ≠ me, hP(h1), ohP(o)(h0), r ∈ h0, h0 ⊥ h1, ¬ ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ((!(me, hO, ohP(o)) * hP))(h0 ∪ h1)

Rule argument

Additional information

Proof step 15 from proof ex-shift-o (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r

Proof step 16 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  ((!(me, hO, ohP(o)) * hP))(h), ¬ ((!(me, hO, _∃ ohP) * hP))(h) ⊦

Rule argument

Additional information

Proof step 17 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

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

• Proof step 19:

  ¬ ((!(me, hO, _∃ ohP) * hP))(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ !(me, hO, ohP(o))(h0) ∧ hP(h1) ⊦

Rule argument

Additional information

Used terms: h, hP, !(me, hO, ohP(o))

Proof step 18 from proof ex-shift-o (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  !(me, hO, ohP(o))(h0), hP(h1), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 20 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

  ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

• Proof step 22:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

Additional information

Used terms: h0, ohP(o), hO, me

Proof step 21 from proof ex-shift-o (previous step) (back to tree)

Lemma owned

Proof step 22 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

Additional information

Proof step 23 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 24:

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

• Proof step 25:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ !(me, hO, _∃ ohP)(h0) ∧ hP(h1) ∧ h0 ⊥ h1

Rule argument

Additional information

Used terms: h1, h0, hP, !(me, hO, _∃ ohP)

Proof step 24 from proof ex-shift-o (previous step) (back to tree)

A spec-lemma

Proof step 25 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ !(me, hO, _∃ ohP)(h0), ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

Additional information

Proof step 26 from proof ex-shift-o (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 27:

  ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

• Proof step 28:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me, ¬ (_∃ ohP)(h0) ∨ (∃ r. ¬(r ∈ h0 → hO(r) = me)) ⊦

Rule argument

Additional information

Used terms: h0, _∃ ohP, hO, me

Proof step 27 from proof ex-shift-o (previous step) (back to tree)

Lemma owned

Proof step 28 from proof ex-shift-o (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me, ¬(∃ o. ohP(o)(h0)) ∨ ¬(r ∈ h0 → hO(r) = me) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ oP)(h) ↔ (∃ o. oP(o)(h)); (s)

Proof step 29 from proof ex-shift-o (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 30:

  hP(h1), ohP(o)(h0), h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. ohP(o)(h0)

• Proof step 31:

  hO(r) ≠ me, hP(h1), ohP(o)(h0), r ∈ h0, h0 ⊥ h1, ¬ ((!(me, hO, _∃ ohP) * hP))(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

Additional information

Proof step 30 from proof ex-shift-o (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: o

Proof step 31 from proof ex-shift-o (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r