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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  h1 ∪ h0 ≠ h0 ∪ h1, h0 ⊥ h1 ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  h1 ∪ h0 ≠ h0 ∪ h1, h0 ⊥ h1 ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

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

• Proof step 5:

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

Rule argument

Additional information

Used terms: h0 ∪ h1, h1 ∪ h0

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

Lemma ext

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  h0 ⊥ h1 ⊦ (r ∈ h1 ∪ h0 ↔ r ∈ h0 ∨ r ∈ h1) ∧ (r ∈ h1 ∪ h0 → (h1 ∪ h0)[r] = (h0 ∪ h1)[r])

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

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

• Proof step 8:

  h0 ⊥ h1, h1 ⊥ h0 ⊦ (r ∈ h1 ∨ r ∈ h0 ↔ r ∈ h0 ∨ r ∈ h1) ∧ (r ∈ h1 ∨ r ∈ h0 → (h1 ∪ h0)[r] = (h0 ∪ h1)[r])

Rule argument

Additional information

Used simplifier rules: disj-com-rw: ⊦ h0 ⊥ h1 → h1 ⊥ h0; Used terms: h0, h1, r

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

Lemma in-uni

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  (h1 ∪ h0)[r] ≠ (h0 ∪ h1)[r], h0 ⊥ h1, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: disj-com-rw: ⊦ h0 ⊥ h1 → h1 ⊥ h0;

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  (r ∈ h1 ⊃ h1[r];h0[r]) ≠ (h0 ∪ h1)[r], r ∈ h1 ∪ h0, h0 ⊥ h1, h1 ⊥ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); disj-com-rw: ⊦ h0 ⊥ h1 → h1 ⊥ h0; Used terms: r, h0, h1

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

Lemma at-uni

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  (r ∈ h1 ⊃ h1[r];h0[r]) ≠ (h0 ∪ h1)[r], r ∈ h1 ∪ h0, h0 ⊥ h1, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: disj-com-rw: ⊦ h0 ⊥ h1 → h1 ⊥ h0;

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); disj-l: ⊦ r ∈ h0 ∧ h0 ⊥ h1 → (r ∈ h1 ↔ false); disj-r: ⊦ r ∈ h1 ∧ h0 ⊥ h1 → (r ∈ h0 ↔ false); Used terms: r, h1, h0

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

Lemma at-uni