Proof step 1 from proof subset-uni (conclusion) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 2:

  h0 ⊆ h1 ⊦ ∃ h. h0 ∪ h = h1 ∧ h ⊥ h0

Rule argument

Additional information

Used simplifier rules: subset-union2: ⊦ h1 ⊥ h0 → h0 ⊆ h0 ∪ h1;

Proof step 2 from proof subset-uni (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 3:

  h0 ⊆ h1 ⊦ h0 ∪ (h1 \ h0) = h1 ∧ h1 \ h0 ⊥ h0

Rule argument

Additional information

Used terms: h1 \ h0

Proof step 3 from proof subset-uni (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 4:

  h0 ⊆ h1, h1 \ h0 ⊥ h0 ⊦ h0 ∪ (h1 \ h0) = h1 ∧ h1 \ h0 ⊥ h0

• Proof step 18:

  h0 ⊆ h1, ¬ h1 \ h0 ⊥ h0 ⊦ h0 ∪ (h1 \ h0) = h1 ∧ h1 \ h0 ⊥ h0

Rule argument

Additional information

Used formulas: h1 \ h0 ⊥ h0

Proof step 4 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  h0 ∪ (h1 \ h0) ≠ h1, h0 ⊆ h1, h1 \ h0 ⊥ h0 ⊦

Rule argument

Additional information

Proof step 5 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 7:

  (h1 \ h0) ∪ h0 ≠ h1, h0 ⊆ h1, h1 \ h0 ⊥ h0 ⊦

Rule argument

Additional information

Used terms: h1 \ h0, h0

Proof step 6 from proof subset-uni (previous step) (back to tree)

Lemma uni-com

Proof step 7 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ (a ∈ st1 → st1(a) = st2(a)))

• Proof step 9:

  h0 ⊆ h1, h1 \ h0 ⊥ h0, ∃ a. ¬((a ∈ (h1 \ h0) ∪ h0 ↔ a ∈ h1) ∧ (a ∈ (h1 \ h0) ∪ h0 → ((h1 \ h0) ∪ h0)(a) = h1(a))) ⊦

Rule argument

Additional information

Used terms: (h1 \ h0) ∪ h0, h1, h1, (h1 \ h0) ∪ h0

Proof step 8 from proof subset-uni (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  h0 ⊆ h1, h1 \ h0 ⊥ h0 ⊦ (r ∈ h1 \ h0 ∨ r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h1 \ h0 ∨ r ∈ h0 → (r ∈ h1 \ h0 ⊃ h1(r);h0(r)) = h1(r))

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); at-uni: r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); disj-l: ⊦ r ∈ h0 ∧ h0 ⊥ h1 → (r ∈ h1 ↔ false); diff-at: r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r);

Proof step 10 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

• Proof step 12:

  h0 ⊆ h1, h1 \ h0 ⊥ h0 ⊦ (r ∈ h1 ∧ ¬ r ∈ h0 ∨ r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h1 ∧ ¬ r ∈ h0 ∨ r ∈ h0 → (r ∈ h1 ∧ ¬ r ∈ h0 ⊃ h1(r);h0(r)) = h1(r))

Rule argument

Additional information

Used terms: r, h1, h0

Proof step 11 from proof subset-uni (previous step) (back to tree)

Lemma diff-in

Proof step 12 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  h0 ⊆ h1, h1 \ h0 ⊥ h0 ⊦ (r ∈ h0 → r ∈ h1) ∧ (r ∈ h1 ∨ r ∈ h0 → (r ∈ h1 ∧ ¬ r ∈ h0 ⊃ h1(r);h0(r)) = h1(r))

Rule argument

Additional information

Proof step 13 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ st1 ⊆ st2 ↔ (∀ a. a ∈ st1 → a ∈ st2 ∧ st1(a) = st2(a))

• Proof step 15:

  h1 \ h0 ⊥ h0, ∀ a. a ∈ h0 → a ∈ h1 ∧ h0(a) = h1(a) ⊦ (r ∈ h0 → r ∈ h1) ∧ (r ∈ h1 ∨ r ∈ h0 → (r ∈ h1 ∧ ¬ r ∈ h0 ⊃ h1(r);h0(r)) = h1(r))

Rule argument

Additional information

Used terms: h0, h1

Proof step 14 from proof subset-uni (previous step) (back to tree)

A spec-lemma

Proof step 15 from proof subset-uni (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 16:

  h1 \ h0 ⊥ h0, ∀ a. a ∈ h0 → a ∈ h1 ∧ h0(a) = h1(a), r ∈ h0 → r ∈ h1 ∧ h0(r) = h1(r) ⊦ (r ∈ h0 → r ∈ h1) ∧ (r ∈ h1 ∨ r ∈ h0 → (r ∈ h1 ∧ ¬ r ∈ h0 ⊃ h1(r);h0(r)) = h1(r))

Rule argument

Additional information

Used terms: r

Proof step 16 from proof subset-uni (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 17:

  (r ∈ h1 ∧ ¬ r ∈ h0 ⊃ h1(r);h0(r)) ≠ h1(r), h1 \ h0 ⊥ h0, ∀ a. a ∈ h0 → a ∈ h1 ∧ h0(a) = h1(a), r ∈ h1 ∨ r ∈ h0, r ∈ h0 → r ∈ h1 ∧ h0(r) = h1(r) ⊦

Rule argument

Additional information

Proof step 17 from proof subset-uni (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

Proof step 18 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  h0 ⊆ h1, ¬ h1 \ h0 ⊥ h0 ⊦

Rule argument

Additional information

Proof step 19 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

  ⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1)

Rule argument

Additional information

Used simplifier rules: diff-in: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1; Used terms: h0, h1 \ h0

Proof step 20 from proof subset-uni (previous step) (back to tree)

Lemma disj