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

• Proof step 36:

  ¬ h0 ⊆ h1, ∃ h. h0 ∪ h = h1 ∧ h ⊥ h0 ⊦

Rule argument

Additional information

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

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2; 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 ⊦

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2;

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 simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2; 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:

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

• Proof step 9:

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

Rule argument

Additional information

Used terms: 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 ⊦ (r ∈ (h1 \ h0) ∪ h0 ↔ r ∈ h1) ∧ (r ∈ (h1 \ h0) ∪ h0 → ((h1 \ h0) ∪ h0)(r) = h1(r))

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2;

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:

  ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1)

• Proof step 12:

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

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2; Used terms: r, h1 \ h0, h0

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

Lemma in-uni

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 ⊦ (r ∈ h1 \ h0 ∨ r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h1 \ h0 ∨ r ∈ h0 → ((h1 \ h0) ∪ h0)(r) = h1(r))

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2;

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:

  ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

• Proof step 15:

  h0 ⊆ h1 ⊦ (r ∈ h1 ∧ ¬ r ∈ h0 ∨ r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h1 ∧ ¬ r ∈ h0 ∨ r ∈ h0 → ((h1 \ h0) ∪ h0)(r) = h1(r))

Rule argument

Additional information

Used terms: r, h1, h0

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

Lemma diff-in

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  ((h1 \ h0) ∪ h0)(r) ≠ h1(r), h0 ⊆ h1, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: subset-in: ⊦ h0 ⊆ h1 ∧ r ∈ h0 → r ∈ h1;

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r))

• Proof step 18:

  ((h1 \ h0) ∪ h0)(r) ≠ h1(r), h0 ⊆ h1, r ∈ h1 ∨ r ∈ h0 ⊦ r ∈ (h1 \ h0) ∪ h0 ∧ h1 \ h0 ⊥ h0

• Proof step 21:

  (r ∈ h1 \ h0 ⊃ (h1 \ h0)(r);h0(r)) ≠ h1(r), h0 ⊆ h1, h1 \ h0 ⊥ h0, r ∈ (h1 \ h0) ∪ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used terms: h1 \ h0, h0, r

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

Lemma at-uni

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:

  ((h1 \ h0) ∪ h0)(r) ≠ h1(r), h0 ⊆ h1, r ∈ h1, ¬ r ∈ h1 \ h0, ¬ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); disj-diff-same: ⊦ h1 \ h2 ⊥ h2;

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:

  ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Rule argument

Additional information

Used terms: r, h1, h0

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

Lemma diff-in

Proof step 21 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  (r ∈ h1 \ h0 ⊃ h1(r);h0(r)) ≠ h1(r), h0 ⊆ h1, r ∈ (h1 \ h0) ∪ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: disj-l: ⊦ r ∈ h0 ∧ h0 ⊥ h1 → ¬ r ∈ h1; diff-in: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1; diff-at: r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r); disj-diff-same: ⊦ h1 \ h2 ⊥ h2;

Proof step 22 from proof subset-uni (previous step) (back to tree)

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

• Proof step 23:

  (r ∈ h1 \ h0 ⊃ h1(r);h0(r)) ≠ h1(r), h0 ⊆ h1, r ∈ (h1 \ h0) ∪ h0, ¬ r ∈ h1 \ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used formulas: r ∈ h1 \ h0

Proof step 23 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ (h1 \ h0) ∪ h0, ¬ r ∈ h1 \ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Proof step 24 from proof subset-uni (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 25:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ (h1 \ h0) ∪ h0, r ∈ h1, ¬ r ∈ h1 \ h0 ⊦

• Proof step 33:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ (h1 \ h0) ∪ h0, r ∈ h0, ¬ r ∈ h1 \ h0 ⊦

Rule argument

Additional information

Proof step 25 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

• Proof step 27:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ h1, r ∈ (h1 \ h0) ∪ h0, ¬ r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used terms: r, h1, h0

Proof step 26 from proof subset-uni (previous step) (back to tree)

Lemma diff-in

Proof step 27 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ (h1 \ h0) ∪ h0, r ∈ h1, r ∈ h0 ⊦

Rule argument

Additional information

Proof step 28 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1)

• Proof step 30:

  h0(r) ≠ h1(r), h0 ⊆ h1, h1 \ h0 ⊥ h0, r ∈ h0, r ∈ h1, r ∈ h1 \ h0 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2; Used terms: r, h1 \ h0, h0

Proof step 29 from proof subset-uni (previous step) (back to tree)

Lemma in-uni

Proof step 30 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ h1, r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: disj-diff-same: ⊦ h1 \ h2 ⊥ h2;

Proof step 31 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

  ⊦ h0 ⊆ h1 ∧ r ∈ h0 → h1(r) = h0(r)

Rule argument

Additional information

Used terms: h0, h1, r

Proof step 32 from proof subset-uni (previous step) (back to tree)

Lemma subset-at

Proof step 33 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  h0(r) ≠ h1(r), h0 ⊆ h1, r ∈ h0, r ∈ (h1 \ h0) ∪ h0, ¬ r ∈ h1 \ h0 ⊦

Rule argument

Additional information

Proof step 34 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ h0 ⊆ h1 ∧ r ∈ h0 → h1(r) = h0(r)

Rule argument

Additional information

Used terms: h0, r, h1

Proof step 35 from proof subset-uni (previous step) (back to tree)

Lemma subset-at

Proof step 36 from proof subset-uni (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  h ⊥ h0, ¬ h0 ⊆ h0 ∪ h ⊦

Rule argument

Additional information

Proof step 37 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 38:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 39:

  h ⊥ h0, ¬ h0 ⊆ h ∪ h0 ⊦

Rule argument

Additional information

Used terms: h, h0

Proof step 38 from proof subset-uni (previous step) (back to tree)

Lemma uni-com

Proof step 39 from proof subset-uni (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 40:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → 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-r: ⊦ r ∈ h1 ∧ h0 ⊥ h1 → ¬ r ∈ h0; Used terms: h ∪ h0, h0

Proof step 40 from proof subset-uni (previous step) (back to tree)

Lemma subset