Proof step 1 from proof union-diff-distr (conclusion) (back to tree)

h1 ⊥ h2 ⊦ (h1 ∪ h2) \ h = (h1 \ h) ∪ (h2 \ h)

System applied rule normalize

And obtained the following new goals

Proof step 2:
(h1 ∪ h2) \ h ≠ (h1 \ h) ∪ (h2 \ h), h1 ⊥ h2 ⊦

Rule argument

Additional information

Proof step 2 from proof union-diff-distr (previous step) (back to tree)

(h1 ∪ h2) \ h ≠ (h1 \ h) ∪ (h2 \ h), h1 ⊥ h2 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))
Proof step 4:
h1 ⊥ h2, ∃ r. ¬((r ∈ (h1 ∪ h2) \ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ (r ∈ (h1 ∪ h2) \ h → ((h1 ∪ h2) \ h)(r) = ((h1 \ h) ∪ (h2 \ h))(r))) ⊦

Rule argument

spec: heap, name:ext, seq: ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r))), subst:[h1, h0] → [(h1 \ h) ∪ (h2 \ h), (h1 ∪ h2) \ h]; [A] → [A], lhs:(h1 ∪ h2) \ h = (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

Used terms: (h1 \ h) ∪ (h2 \ h), (h1 ∪ h2) \ h

Proof step 3 from proof union-diff-distr (previous step) (back to tree)

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

A spec-lemma

Proof step 4 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2, ∃ r. ¬((r ∈ (h1 ∪ h2) \ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ (r ∈ (h1 ∪ h2) \ h → ((h1 ∪ h2) \ h)(r) = ((h1 \ h) ∪ (h2 \ h))(r))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
h1 ⊥ h2 ⊦ (r ∈ (h1 ∪ h2) \ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ (r ∈ (h1 ∪ h2) \ h → ((h1 ∪ h2) \ h)(r) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

Additional information

Proof step 5 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ (r ∈ (h1 ∪ h2) \ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ (r ∈ (h1 ∪ h2) \ h → ((h1 ∪ h2) \ h)(r) = ((h1 \ h) ∪ (h2 \ h))(r))

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1
Proof step 7:
h1 ⊥ h2 ⊦ (r ∈ h1 ∪ h2 ∧ ¬ r ∈ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ (r ∈ h1 ∪ h2 ∧ ¬ r ∈ h → ((h1 ∪ h2) \ h)(r) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h1 ∪ h2, h]; [A] → [A], lhs:r ∈ (h1 ∪ h2) \ h, <all paths>

Additional information

Used terms: r, h1 ∪ h2, h

Proof step 6 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof step 7 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ (r ∈ h1 ∪ h2 ∧ ¬ r ∈ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ (r ∈ h1 ∪ h2 ∧ ¬ r ∈ h → ((h1 ∪ h2) \ h)(r) = ((h1 \ h) ∪ (h2 \ h))(r))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

Additional information

Used simplifier rules: diff-at: r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r); 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));

Proof step 8 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ (h1 \ h) ∪ (h2 \ h)) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1)
Proof step 10:
h1 ⊥ h2, h1 \ h ⊥ h2 \ h ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 \ h ∨ r ∈ h2 \ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

spec: <toplevel>, name:in-uni, seq: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1), subst:[r, h0, h1] → [r, h1 \ h, h2 \ h]; [A] → [A], lhs:r ∈ (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

Used simplifier rules: disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3; Used terms: r, h1 \ h, h2 \ h

Proof step 9 from proof union-diff-distr (previous step) (back to tree)

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

Lemma in-uni

Proof step 10 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2, h1 \ h ⊥ h2 \ h ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 \ h ∨ r ∈ h2 \ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 \ h ∨ r ∈ h2 \ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

Additional information

Used simplifier rules: disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3;

Proof step 11 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 \ h ∨ r ∈ h2 \ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1
Proof step 13:
h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 \ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h1, h]; [A] → [A], lhs:r ∈ h1 \ h, <all paths>

Additional information

Used terms: r, h1, h

Proof step 12 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof step 13 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 \ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

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:
h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h2, h]; [A] → [A], lhs:r ∈ h2 \ h, <all paths>

Additional information

Used terms: r, h2, h

Proof step 14 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof step 15 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h ↔ r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
h1 ⊥ h2 ⊦ (r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h → (r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

Additional information

Proof step 16 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2 ⊦ (r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h → (r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

User applied rule cut formula

And obtained the following new goals

Proof step 17:
h1 ⊥ h2, r ∈ h1 ⊦ (r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h → (r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))
Proof step 30:
h1 ⊥ h2, ¬ r ∈ h1 ⊦ (r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h → (r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Rule argument

fma: r ∈ h1

Additional information

Used formulas: r ∈ h1

Proof step 17 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2, r ∈ h1 ⊦ (r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h → (r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
h1(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h1, ¬ r ∈ h ⊦

Rule argument

Additional information

Used simplifier rules: disj-l: ⊦ r ∈ h0 ∧ h0 ⊥ h1 → ¬ r ∈ h1;

Proof step 18 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h1, ¬ r ∈ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 19:
r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r))
Proof step 20:
h1(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h1, ¬ r ∈ h ⊦ r ∈ (h1 \ h) ∪ (h2 \ h) ∧ h1 \ h ⊥ h2 \ h
Proof step 23:
h1(r) ≠ (r ∈ h1 \ h ⊃ (h1 \ h)(r);(h2 \ h)(r)), h1 ⊥ h2, h1 \ h ⊥ h2 \ h, r ∈ h1, r ∈ (h1 \ h) ∪ (h2 \ h), ¬ r ∈ h ⊦

Rule argument

spec: <toplevel>, name:at-uni, seq:r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)), subst:[h0, h1, r] → [h1 \ h, h2 \ h, r]; [A] → [A], lhs:((h1 \ h) ∪ (h2 \ h))(r), <all paths>

Additional information

Used terms: h1 \ h, h2 \ h, r

Proof step 19 from proof union-diff-distr (previous step) (back to tree)

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

Lemma at-uni

Proof step 20 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h1, ¬ r ∈ h ⊦ r ∈ (h1 \ h) ∪ (h2 \ h) ∧ h1 \ h ⊥ h2 \ h

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
h1(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h1, ¬ r ∈ h, ¬ r ∈ h1 \ h, ¬ r ∈ h2 \ h ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3;

Proof step 21 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h1, ¬ r ∈ h, ¬ r ∈ h1 \ h, ¬ r ∈ h2 \ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h1, h]; [A] → [A], lhs:r ∈ h1 \ h, <all paths>

Additional information

Used terms: r, h1, h

Proof step 22 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof step 23 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ (r ∈ h1 \ h ⊃ (h1 \ h)(r);(h2 \ h)(r)), h1 ⊥ h2, h1 \ h ⊥ h2 \ h, r ∈ h1, r ∈ (h1 \ h) ∪ (h2 \ h), ¬ r ∈ h ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
h1(r) ≠ (r ∈ h1 \ h ⊃ h1(r);(h2 \ h)(r)), h1 ⊥ h2, r ∈ (h1 \ h) ∪ (h2 \ h), r ∈ h1, ¬ r ∈ h ⊦

Rule argument

Additional information

Used simplifier rules: diff-at: r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r); disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3;

Proof step 24 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ (r ∈ h1 \ h ⊃ h1(r);(h2 \ h)(r)), h1 ⊥ h2, r ∈ (h1 \ h) ∪ (h2 \ h), r ∈ h1, ¬ r ∈ h ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 25:
h1(r) ≠ (h2 \ h)(r), h1 ⊥ h2, r ∈ h1, r ∈ (h1 \ h) ∪ (h2 \ h), ¬ r ∈ h1 \ h, ¬ r ∈ h ⊦

Rule argument

pos: left 1

Additional information

Proof step 25 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ (h2 \ h)(r), h1 ⊥ h2, r ∈ h1, r ∈ (h1 \ h) ∪ (h2 \ h), ¬ r ∈ h1 \ h, ¬ r ∈ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 26:
r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r)
Proof step 27:
h1(r) ≠ (h2 \ h)(r), h1 ⊥ h2, r ∈ (h1 \ h) ∪ (h2 \ h), r ∈ h1, ¬ r ∈ h1 \ h, ¬ r ∈ h ⊦ r ∈ h2 ∧ ¬ r ∈ h

Rule argument

spec: <toplevel>, name:diff-at, seq:r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r), subst:[h0, h1, r] → [h2, h, r]; [A] → [A], lhs:(h2 \ h)(r), <all paths>

Additional information

Used simplifier rules: disj-l: ⊦ r ∈ h0 ∧ h0 ⊥ h1 → ¬ r ∈ h1; Used terms: h2, h, r

Proof step 26 from proof union-diff-distr (previous step) (back to tree)

r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r)

Lemma diff-at

Proof step 27 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ (h2 \ h)(r), h1 ⊥ h2, r ∈ (h1 \ h) ∪ (h2 \ h), r ∈ h1, ¬ r ∈ h1 \ h, ¬ r ∈ h ⊦ r ∈ h2 ∧ ¬ r ∈ h

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
h1(r) ≠ (h2 \ h)(r), h1 ⊥ h2, r ∈ h1, r ∈ (h1 \ h) ∪ (h2 \ h), ¬ r ∈ h1 \ h, ¬ r ∈ h, ¬ r ∈ h2 ⊦

Rule argument

Additional information

Proof step 28 from proof union-diff-distr (previous step) (back to tree)

h1(r) ≠ (h2 \ h)(r), h1 ⊥ h2, r ∈ h1, r ∈ (h1 \ h) ∪ (h2 \ h), ¬ r ∈ h1 \ h, ¬ r ∈ h, ¬ r ∈ h2 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h1, h]; [A] → [A], lhs:r ∈ h1 \ h, <all paths>

Additional information

Used terms: r, h1, h

Proof step 29 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof step 30 from proof union-diff-distr (previous step) (back to tree)

h1 ⊥ h2, ¬ r ∈ h1 ⊦ (r ∈ h1 ∧ ¬ r ∈ h ∨ r ∈ h2 ∧ ¬ r ∈ h → (r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h) ∧ ((r ∈ h1 ∨ r ∈ h2) ∧ ¬ r ∈ h → (r ∈ h1 ⊃ h1(r);h2(r)) = ((h1 \ h) ∪ (h2 \ h))(r))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h ⊦

Rule argument

Additional information

Proof step 31 from proof union-diff-distr (previous step) (back to tree)

h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 32:
r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r))
Proof step 33:
h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h ⊦ r ∈ (h1 \ h) ∪ (h2 \ h) ∧ h1 \ h ⊥ h2 \ h

Rule argument

Additional information

Used simplifier rules: diff-at: r ∈ h0, ¬ r ∈ h1 ⊦ (h0 \ h1)(r) = h0(r); disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3; diff-in: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1; disj-r: ⊦ r ∈ h1 ∧ h0 ⊥ h1 → ¬ r ∈ h0; Used terms: h1 \ h, h2 \ h, r

Proof step 32 from proof union-diff-distr (previous step) (back to tree)

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

Lemma at-uni

Proof step 33 from proof union-diff-distr (previous step) (back to tree)

h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h ⊦ r ∈ (h1 \ h) ∪ (h2 \ h) ∧ h1 \ h ⊥ h2 \ h

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h, ¬ r ∈ h1 \ h, ¬ r ∈ h2 \ h ⊦

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3;

Proof step 34 from proof union-diff-distr (previous step) (back to tree)

h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h, ¬ r ∈ h1 \ h, ¬ r ∈ h2 \ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 35:
⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1
Proof step 36:
h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h, ¬ r ∈ h2 \ h, ¬ r ∈ h1 ∨ r ∈ h ⊦

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h1, h]; [A] → [A], lhs:r ∈ h1 \ h, <all paths>

Additional information

Used terms: r, h1, h

Proof step 35 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof step 36 from proof union-diff-distr (previous step) (back to tree)

h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h, ¬ r ∈ h2 \ h, ¬ r ∈ h1 ∨ r ∈ h ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h, ¬ r ∈ h2 \ h ⊦

Rule argument

Additional information

Proof step 37 from proof union-diff-distr (previous step) (back to tree)

h2(r) ≠ ((h1 \ h) ∪ (h2 \ h))(r), h1 ⊥ h2, r ∈ h2, ¬ r ∈ h1, ¬ r ∈ h, ¬ r ∈ h2 \ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 38:
⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Rule argument

spec: <toplevel>, name:diff-in, seq: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1, subst:[r, h0, h1] → [r, h2, h]; [A] → [A], lhs:r ∈ h2 \ h, <all paths>

Additional information

Used terms: r, h2, h

Proof step 38 from proof union-diff-distr (previous step) (back to tree)

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

Proof Tree for union-diff-distr (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.