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

⊦ s ⊆ s0 ↔ (∃ s1. s ∪ s1 = s0 ∧ s ∩ s1 = ∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
s ⊆ s0 ⊦ ∃ s1. s ∪ s1 = s0 ∧ s ∩ s1 = ∅

Rule argument

Additional information

Used simplifier rules: unsub: ⊦ s ⊆ s ∪ s0; una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; unc: ⊦ s ∪ s0 = s0 ∪ s;

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

s ⊆ s0 ⊦ ∃ s1. s ∪ s1 = s0 ∧ s ∩ s1 = ∅

User applied rule structural induction

And obtained the following new goals

Proof step 3:
s ⊆ s1 ++ elem, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

indvar: s0

Additional information

Used simplifier rules: une-01: ⊦ s1 ∪ s2 = ∅ ↔ s1 = ∅ ∧ s2 = ∅; una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; unc: ⊦ s ∪ s0 = s0 ∪ s; int-e: ⊦ s ∩ ∅ = ∅; int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; int-c: ⊦ s ∩ s0 = s0 ∩ s; set generated by ∅, ++;; ⊦ s ⊆ ∅ ↔ s = ∅; (s)

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

s ⊆ s1 ++ elem, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ s → s ++ a = s
Proof step 5:
s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅
Proof step 24:
elem ∈ s1, s ⊆ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Rule argument

spec: set-basic, name:eq, seq: ⊦ a ∈ s → s ++ a = s, subst:[a, s] → [elem, s1], <all paths>

Additional information

Used terms: elem, s1

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

⊦ a ∈ s → s ++ a = s

A spec-lemma

Proof step 5 from proof subset-is-union (previous step) (back to tree)

s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

Additional information

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

s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0)
Proof step 8:
elem ∈ s, s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅
Proof step 20:
s ⊆ s1, ¬ elem ∈ s, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

spec: set-basic, name:sub-insert, seq: ⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0), subst:[a, s0, s] → [elem, s1, s], <all paths>

Additional information

Used terms: elem, s1, s

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

⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0)

A spec-lemma

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

elem ∈ s, s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# s ≠ 0, s ≠ ∅, elem ∈ s, s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b); (s) ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic

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

# s ≠ 0, s ≠ ∅, elem ∈ s, s ⊆ s1 ++ elem, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

User applied rule all left

And obtained the following new goals

Proof step 10:
# s ≠ 0, s ≠ ∅, elem ∈ s, s ⊆ s1 ++ elem, ¬ elem ∈ s1, s -- elem ⊆ s1 → (∃ s2. (s -- elem) ∪ s2 = s1 ∧ (s -- elem) ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

pos: left 6 terms: s -- elem abort: F discard: T computed: F

Additional information

Used terms: s -- elem

Proof step 10 from proof subset-is-union (previous step) (back to tree)

# s ≠ 0, s ≠ ∅, elem ∈ s, s ⊆ s1 ++ elem, ¬ elem ∈ s1, s -- elem ⊆ s1 → (∃ s2. (s -- elem) ∪ s2 = s1 ∧ (s -- elem) ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(s -- elem) ∩ s2 = ∅, # s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦ ∃ s0. s ∪ s0 = s2 ∪ s ∧ s ∩ s0 = ∅

Rule argument

Additional information

Used simplifier rules: insdel-sub: ⊦ ¬ s -- a ⊆ s0 ∧ a ∈ s → ¬ s ⊆ s0 ++ a; Union: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2; Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; unsub: ⊦ s ⊆ s ∪ s0; undel-03: ⊦ ¬ a ∈ s → s ∪ (s0 -- a) = (s ∪ s0) -- a; delete-insert: ⊦ (s -- a) ++ a = s ++ a; uninsert: ⊦ a ∈ s → (s ∪ s0) ++ a = s ∪ s0; una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; unc: ⊦ s ∪ s0 = s0 ∪ s;

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

(s -- elem) ∩ s2 = ∅, # s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦ ∃ s0. s ∪ s0 = s2 ∪ s ∧ s ∩ s0 = ∅

User applied rule exists right

And obtained the following new goals

Proof step 12:
(s -- elem) ∩ s2 = ∅, # s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦ s ∪ s2 = s2 ∪ s ∧ s ∩ s2 = ∅

Rule argument

pos: right 1 terms: s2 abort: F discard: T computed: F

Additional information

Used terms: s2

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

(s -- elem) ∩ s2 = ∅, # s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦ s ∪ s2 = s2 ∪ s ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(s -- elem) ∩ s2 = ∅, #(s ∩ s2) ≠ 0, # s ≠ 0, s ≠ ∅, s ∩ s2 ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦

Rule argument

Additional information

Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s; size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic

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

(s -- elem) ∩ s2 = ∅, #(s ∩ s2) ≠ 0, # s ≠ 0, s ≠ ∅, s ∩ s2 ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ s ∩ (s0 -- a) = (s ∩ s0) -- a
Proof step 15:
(s2 ∩ s) -- elem = ∅, #(s ∩ s2) ≠ 0, # s ≠ 0, s ≠ ∅, s ∩ s2 ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦

Rule argument

spec: <toplevel>, name:int-del-rew, seq: ⊦ s ∩ (s0 -- a) = (s ∩ s0) -- a, subst:[s, a, s0] → [s2, elem, s], <all paths>

Additional information

Used simplifier rules: int-c: ⊦ s ∩ s0 = s0 ∩ s; Used terms: s2, elem, s

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

⊦ s ∩ (s0 -- a) = (s ∩ s0) -- a

Lemma int-del-rew

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

(s2 ∩ s) -- elem = ∅, #(s ∩ s2) ≠ 0, # s ≠ 0, s ≠ ∅, s ∩ s2 ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
s2 ∩ s = ∅ ++ elem, # s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦

Rule argument

Additional information

Used simplifier rules: int-c: ⊦ s ∩ s0 = s0 ∩ s; del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; ⊦ #(∅ ++ a) = 1; (s) ⊦ ∅ ≠ s ++ a; (s) size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic

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

s2 ∩ s = ∅ ++ elem, # s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 18:
# s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2, ∀ a. a ∈ s2 ∩ s ↔ a ∈ ∅ ++ elem ⊦

Rule argument

spec: set-basic, name:Extension, seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2), subst:[s2, s1] → [∅ ++ elem, s2 ∩ s], <all paths>

Additional information

Used terms: ∅ ++ elem, s2 ∩ s

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

⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

A spec-lemma

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

# s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2, ∀ a. a ∈ s2 ∩ s ↔ a ∈ ∅ ++ elem ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2, ∀ a. a ∈ s2 ∧ a ∈ s ↔ a = elem ⊦

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2; int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; int-c: ⊦ s ∩ s0 = s0 ∩ s; ⊦ a ∈ ∅ ++ b ↔ a = b; (s)

Proof step 19 from proof subset-is-union (previous step) (back to tree)

# s ≠ 0, s ≠ ∅, elem ∈ s, ¬ elem ∈ s2, ∀ a. a ∈ s2 ∧ a ∈ s ↔ a = elem ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: elem

Additional information

Used terms: elem

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

s ⊆ s1, ¬ elem ∈ s, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
s ⊆ s1, ¬ elem ∈ s, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

Additional information

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

s ⊆ s1, ¬ elem ∈ s, ¬ elem ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

User applied rule all left

And obtained the following new goals

Proof step 22:
s ⊆ s1, ¬ elem ∈ s, ¬ elem ∈ s1, s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅), ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Rule argument

pos: left 4 terms: s abort: F discard: F computed: T

Additional information

Used terms: s

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

s ⊆ s1, ¬ elem ∈ s, ¬ elem ∈ s1, s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅), ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ++ elem ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
s ∩ s2 = ∅, ¬ elem ∈ s, ¬ elem ∈ s2, ∀ s0. s0 ⊆ s ∪ s2 → (∃ s3. s0 ∪ s3 = s ∪ s2 ∧ s0 ∩ s3 = ∅) ⊦ ∃ s0. s ∪ s0 = (s ∪ s2) ++ elem ∧ s ∩ s0 = ∅

Rule argument

Additional information

Used simplifier rules: unsub: ⊦ s ⊆ s ∪ s0; Union: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2; una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; unc: ⊦ s ∪ s0 = s0 ∪ s;

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

s ∩ s2 = ∅, ¬ elem ∈ s, ¬ elem ∈ s2, ∀ s0. s0 ⊆ s ∪ s2 → (∃ s3. s0 ∪ s3 = s ∪ s2 ∧ s0 ∩ s3 = ∅) ⊦ ∃ s0. s ∪ s0 = (s ∪ s2) ++ elem ∧ s ∩ s0 = ∅

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: s2 ++ elem abort: F discard: T computed: F

Additional information

Used simplifier rules: unadd: ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a; una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; unc: ⊦ s ∪ s0 = s0 ∪ s; int-empty: ⊦ s ∩ s0 = ∅ ∧ ¬ a ∈ s → s ∩ (s0 ++ a) = ∅; int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; int-c: ⊦ s ∩ s0 = s0 ∩ s; Used terms: s2 ++ elem

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

elem ∈ s1, s ⊆ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
# s1 ≠ 0, s1 ≠ ∅, elem ∈ s1, s ⊆ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic

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

# s1 ≠ 0, s1 ≠ ∅, elem ∈ s1, s ⊆ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: s

Additional information

Used terms: s

Proof Tree for subset-is-union (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.