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

⊦ h0 ⊆ h1 ↔ (∃ h. h0 ∪ h = h1 ∧ h ⊥ h0)

User applied rule case distinction

And obtained the following new goals

Proof step 2:
h0 ⊆ h1 ⊦ ∃ h. h0 ∪ h = h1 ∧ h ⊥ h0
Proof step 24:
¬ h0 ⊆ h1, ∃ h. h0 ∪ h = h1 ∧ h ⊥ h0 ⊦

Rule argument

pos: right 1

Additional information

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

h0 ⊆ h1 ⊦ ∃ h. h0 ∪ h = h1 ∧ h ⊥ h0

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

pos: right 1 terms: h1 \ h0 abort: F discard: T computed: F

Additional information

Used terms: h1 \ h0

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

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

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

fma: h1 \ h0 ⊥ h0

Additional information

Used formulas: h1 \ h0 ⊥ h0

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

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

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)

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

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

spec: <toplevel>, name: uni-com, seq: ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1, subst:[h0, h1] → [h1 \ h0, h0], <all paths>

Additional information

Used terms: h1 \ h0, h0

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

⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

Lemma uni-com

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

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

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

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

Additional information

Used terms: h1, (h1 \ h0) ∪ h0

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

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

Lemma ext

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

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

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)

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))

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

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

Additional information

Used terms: r, h1, h0

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

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

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

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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(r ∈ h1 ∧ ¬ r ∈ h0 ⊃ h1(r);h0(r)) ≠ h1(r), h0 ⊆ h1, h1 \ h0 ⊥ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

Additional information

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

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

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
h0(r) ≠ h1(r), h0 ⊆ h1, h1 \ h0 ⊥ h0, ¬ r ∈ h1 ∨ r ∈ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

pos: left 1

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

Proof step 15:
h0(r) ≠ h1(r), h0 ⊆ h1, h1 \ h0 ⊥ h0, r ∈ h0, r ∈ h1 ∨ r ∈ h0 ⊦

Rule argument

pos: left 4

Additional information

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

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
h0(r) ≠ h1(r), h0 ⊆ h1, h1 \ h0 ⊥ h0, r ∈ h0 ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ h0 ⊆ h1 ∧ r ∈ h0 → h1(r) = h0(r)

Rule argument

spec: <toplevel>, name: subset-at, seq: ⊦ h0 ⊆ h1 ∧ r ∈ h0 → h1(r) = h0(r), subst:[h0, r, h1] → [h0, r, h1], <all paths>

Additional information

Used terms: h0, r, h1

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

⊦ h0 ⊆ h1 ∧ r ∈ h0 → h1(r) = h0(r)

Lemma subset-at

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

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

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)

h0 ⊆ h1, ¬ h1 \ h0 ⊥ h0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1)
Proof step 21:
h0 ⊆ h1, ∃ r. ¬(¬ r ∈ h1 \ h0 ∨ ¬ r ∈ h0) ⊦

Rule argument

spec: <toplevel>, name: disj, seq: ⊦ h0 ⊥ h1 ↔ (∀ r. ¬ r ∈ h0 ∨ ¬ r ∈ h1), subst:[h1, h0] → [h0, h1 \ h0], <all paths>

Additional information

Used terms: h0, h1 \ h0

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

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

Lemma disj

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

h0 ⊆ h1, ∃ r. ¬(¬ r ∈ h1 \ h0 ∨ ¬ r ∈ h0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
h0 ⊆ h1, r ∈ h1 \ h0, r ∈ h0 ⊦

Rule argument

Additional information

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

h0 ⊆ h1, r ∈ h1 \ h0, r ∈ h0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ 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, h0], <all paths>

Additional information

Used terms: r, h1, h0

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

⊦ r ∈ h0 \ h1 ↔ r ∈ h0 ∧ ¬ r ∈ h1

Lemma diff-in

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
h ⊥ h0, ¬ h0 ⊆ h0 ∪ h ⊦

Rule argument

Additional information

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

h ⊥ h0, ¬ h0 ⊆ h0 ∪ h ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1
Proof step 27:
h ⊥ h0, ¬ h0 ⊆ h ∪ h0 ⊦

Rule argument

spec: <toplevel>, name: uni-com, seq: ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1, subst:[h0, h1] → [h, h0], <all paths>

Additional information

Used terms: h, h0

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

⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

Lemma uni-com

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

h ⊥ h0, ¬ h0 ⊆ h ∪ h0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

Rule argument

spec: <toplevel>, name: subset, seq: ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r))), subst:[h1, h0] → [h ∪ h0, h0], <all paths>

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 ↔ false); Used terms: h ∪ h0, h0

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

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

Lemma subset

Proof Tree for subset-uni (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.