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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

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

Rule argument

spec: map, name:Extension, seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1(a) = st2(a)), subst:[st2, st1] → [(h1 ∪ h2) \ h, (h1 \ h) ∪ (h2 \ h)]; [A, B] → [A, B], lhs:(h1 ∪ h2) \ h = (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

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

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

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

A spec-lemma

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: map, name:notmemlookup, seq: ⊦ ¬ a ∈ st → st(a) = ∅(a), subst:[st, a] → [(h1 \ h) ∪ (h2 \ h), r]; [A, B] → [A, B], lhs:((h1 \ h) ∪ (h2 \ h))(r), <all paths>

Additional information

Used terms: (h1 \ h) ∪ (h2 \ h), r

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

⊦ ¬ a ∈ st → st(a) = ∅(a)

A spec-lemma

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

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, B] → [A, B], lhs:r ∈ (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

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

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

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

Lemma in-uni

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

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, B] → [A, B], lhs:r ∈ (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

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

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

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

Lemma in-uni

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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

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, B] → [A, B], lhs:r ∈ (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

Used simplifier rules: disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3; 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)); 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-l: ⊦ r ∈ h0 ∧ h0 ⊥ h1 → (r ∈ h1 ↔ false); Used terms: r, h1 \ h, h2 \ h

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

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

Lemma in-uni

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

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, B] → [A, B], lhs:r ∈ (h1 \ h) ∪ (h2 \ h), <all paths>

Additional information

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

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

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

Lemma in-uni

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

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, B] → [A, B], 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 20 from proof union-diff-distr (previous step) (back to tree)

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

Lemma in-uni

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: disj-diffb: h0 ⊥ h1 ⊦ h0 \ h2 ⊥ h1 \ h3; diff-in: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1;

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: map, name:notmemlookup, seq: ⊦ ¬ a ∈ st → st(a) = ∅(a), subst:[st, a] → [(h1 ∪ h2) \ h, r]; [A, B] → [A, B], lhs:((h1 ∪ h2) \ h)(r), <all paths>

Additional information

Used simplifier rules: diff-in: ⊦ 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); diff-in: ⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1; in-uni: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); Used terms: (h1 ∪ h2) \ h, r

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

⊦ ¬ a ∈ st → st(a) = ∅(a)

A spec-lemma

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.

the proof tree