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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬ precise(r.=>_) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ¬ precise(_∃(λ o. r => o)) ⊦

Rule argument

Additional information

Used simplifier rules: mapex: ⊦ r.=>_ = _∃(λ o. r => o);

Proof step 3 from proof precise-mapex (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. h0 ⊆ h ∧ h1 ⊆ h ∧ hP(h0) ∧ hP(h1) → h0 = h1)

• Proof step 5:

  ∃ h, h0, h1. ¬(h0 ⊆ h ∧ h1 ⊆ h ∧ (_∃(λ o. r => o))(h0) ∧ (_∃(λ o. r => o))(h1) → h0 = h1) ⊦

Rule argument

Additional information

Used terms: _∃(λ o. r => o)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  h0 ≠ h1, h0 ⊆ h, h1 ⊆ h, (r => o)(h0), (r => o0)(h1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ bP)(h) ↔ (∃ b. bP(b)(h)); (s)

Proof step 6 from proof precise-mapex (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ (r => o)(h) ↔ h = (∅ ++ r)(r := o) ∧ r ≠ null

• Proof step 8:

  h1 = (∅ ++ r)(r := o0), h0 ≠ h1, r ≠ null, h0 ⊆ h, h1 ⊆ h, (r => o)(h0) ⊦

Rule argument

Additional information

Used terms: r, o0, h1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  h0 ≠ (∅ ++ r)(r := o0), r ≠ null, h0 ⊆ h, (r => o)(h0), (∅ ++ r)(r := o0) ⊆ h ⊦

Rule argument

Additional information

Proof step 9 from proof precise-mapex (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ (r => o)(h) ↔ h = (∅ ++ r)(r := o) ∧ r ≠ null

• Proof step 11:

  h0 = (∅ ++ r)(r := o), h0 ≠ (∅ ++ r)(r := o0), r ≠ null, h0 ⊆ h, (∅ ++ r)(r := o0) ⊆ h ⊦

Rule argument

Additional information

Used terms: r, o, h0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  r ≠ null, (∅ ++ r)(r := o) ≠ (∅ ++ r)(r := o0), (∅ ++ r)(r := o0) ⊆ h, (∅ ++ r)(r := o) ⊆ h ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 14:

  r ≠ null, (∅ ++ r)(r := o) ⊆ h, (∅ ++ r)(r := o0) ⊆ h, ∃ r0. ¬((r0 ∈ (∅ ++ r)(r := o) ↔ r0 ∈ (∅ ++ r)(r := o0)) ∧ (r0 ∈ (∅ ++ r)(r := o) → (∅ ++ r)(r := o)(r0) = (∅ ++ r)(r := o0)(r0))) ⊦

Rule argument

Additional information

Used terms: (∅ ++ r)(r := o), (∅ ++ r)(r := o0)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  o ≠ o0, r ≠ null, (∅ ++ r)(r := o) ⊆ h, (∅ ++ r)(r := o0) ⊆ h ⊦

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) ⊦ ¬ r ∈ ∅; (s) ⊦ h(r0 := o)(r) = (r = r0 ∧ r0 ∈ h ⊃ o;h(r)); (s) ⊦ r ∈ h(r0 := o) ↔ r ∈ h; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 17:

  o ≠ o0, r ≠ null, (∅ ++ r)(r := o) ⊆ h, ∀ r0. (r0 ∈ (∅ ++ r)(r := o0) → r0 ∈ h) ∧ (r0 ∈ (∅ ++ r)(r := o0) → (∅ ++ r)(r := o0)(r0) = h(r0)) ⊦

Rule argument

Additional information

Used terms: (∅ ++ r)(r := o0), h

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  o ≠ o0, r ≠ null, (∅ ++ r)(r := o) ⊆ h, ∀ r0. (r0 = r → r ∈ h) ∧ (r0 = r → o0 = h(r)) ⊦

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) ⊦ ¬ r ∈ ∅; (s) ⊦ h(r0 := o)(r) = (r = r0 ∧ r0 ∈ h ⊃ o;h(r)); (s) ⊦ r ∈ h(r0 := o) ↔ r ∈ h; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 20:

  o ≠ o0, r ≠ null, ∀ r0. (r0 = r → r ∈ h) ∧ (r0 = r → o0 = h(r)), ∀ r0. (r0 ∈ (∅ ++ r)(r := o) → r0 ∈ h) ∧ (r0 ∈ (∅ ++ r)(r := o) → (∅ ++ r)(r := o)(r0) = h(r0)) ⊦

Rule argument

Additional information

Used terms: (∅ ++ r)(r := o), h

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  o ≠ o0, r ≠ null, ∀ r0. (r0 = r → r ∈ h) ∧ (r0 = r → o0 = h(r)), ∀ r0. (r0 = r → r ∈ h) ∧ (r0 = r → o = h(r)) ⊦

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) ⊦ ¬ r ∈ ∅; (s) ⊦ h(r0 := o)(r) = (r = r0 ∧ r0 ∈ h ⊃ o;h(r)); (s) ⊦ r ∈ h(r0 := o) ↔ r ∈ h; (s)

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h → r ≠ null; (s) Used terms: r