Proof step 1 from proof iterate-is-fix (conclusion) (back to tree)

cont(f) ⊦ μ(f)(x) ↔ (∃ n. ((f ^ n))(∅)(x))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
cont(f), μ(f)(x) ⊦ ∃ n. ((f ^ n))(∅)(x)
Proof step 15:
cont(f), ¬ μ(f)(x), ∃ n. ((f ^ n))(∅)(x) ⊦

Rule argument

pos: right 1

Additional information

Proof step 2 from proof iterate-is-fix (previous step) (back to tree)

cont(f), μ(f)(x) ⊦ ∃ n. ((f ^ n))(∅)(x)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ monotone(f) → (μ(f)(x) ↔ (∀ iset. f(iset) ⊆ iset → iset(x)))
Proof step 4:
cont(f), μ(f)(x), ¬ monotone(f) ⊦ ∃ n. ((f ^ n))(∅)(x)
Proof step 7:
cont(f), monotone(f), ∀ iset. f(iset) ⊆ iset → iset(x) ⊦ ∃ n. ((f ^ n))(∅)(x)

Rule argument

spec: monotone-set-fixpoint, name:least-prefixpoint-def, seq: ⊦ monotone(f) → (μ(f)(x) ↔ (∀ iset. f(iset) ⊆ iset → iset(x))), subst:[x, f] → [x, f], <all paths>

Additional information

Used terms: x, f

Proof step 3 from proof iterate-is-fix (previous step) (back to tree)

⊦ monotone(f) → (μ(f)(x) ↔ (∀ iset. f(iset) ⊆ iset → iset(x)))

A spec-lemma

Proof step 4 from proof iterate-is-fix (previous step) (back to tree)

cont(f), μ(f)(x), ¬ monotone(f) ⊦ ∃ n. ((f ^ n))(∅)(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
cont(f), μ(f)(x), ¬ monotone(f) ⊦ ∃ n. ((f ^ n))(∅)(x)

Rule argument

Additional information

Proof step 5 from proof iterate-is-fix (previous step) (back to tree)

cont(f), μ(f)(x), ¬ monotone(f) ⊦ ∃ n. ((f ^ n))(∅)(x)

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
cont(f) ⊦ monotone(f)

Rule argument

spec: <toplevel>, name:cont-monotone, seq:cont(f) ⊦ monotone(f), subst:[f] → [f]

Additional information

Used terms: f

Proof step 6 from proof iterate-is-fix (previous step) (back to tree)

cont(f) ⊦ monotone(f)

Lemma cont-monotone

Proof step 7 from proof iterate-is-fix (previous step) (back to tree)

cont(f), monotone(f), ∀ iset. f(iset) ⊆ iset → iset(x) ⊦ ∃ n. ((f ^ n))(∅)(x)

User applied rule all left

And obtained the following new goals

Proof step 8:
cont(f), monotone(f), f(λ x. ∃ n. ((f ^ n))(∅)(x)) ⊆ (λ x. ∃ n. ((f ^ n))(∅)(x)) → (∃ n. ((f ^ n))(∅)(x)) ⊦ ∃ n. ((f ^ n))(∅)(x)

Rule argument

pos: left 3 terms: λ x. ∃ n. ((f ^ n))(∅)(x) abort: F discard: T computed: F

Additional information

Used terms: λ x. ∃ n. ((f ^ n))(∅)(x)

Proof step 8 from proof iterate-is-fix (previous step) (back to tree)

cont(f), monotone(f), f(λ x. ∃ n. ((f ^ n))(∅)(x)) ⊆ (λ x. ∃ n. ((f ^ n))(∅)(x)) → (∃ n. ((f ^ n))(∅)(x)) ⊦ ∃ n. ((f ^ n))(∅)(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
cont(f), monotone(f) ⊦ ∃ n. ((f ^ n))(∅)(x), f(λ x. ∃ n. ((f ^ n))(∅)(x)) ⊆ (λ x. ∃ n. ((f ^ n))(∅)(x))

Rule argument

Additional information

Proof step 9 from proof iterate-is-fix (previous step) (back to tree)

cont(f), monotone(f) ⊦ ∃ n. ((f ^ n))(∅)(x), f(λ x. ∃ n. ((f ^ n))(∅)(x)) ⊆ (λ x. ∃ n. ((f ^ n))(∅)(x))

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ iset ⊆ iset0 ↔ (∀ x. iset(x) → iset0(x))
Proof step 11:
cont(f), monotone(f) ⊦ ∀ x. f(λ x. ∃ n. ((f ^ n))(∅)(x))(x) → (∃ n. ((f ^ n))(∅)(x)), ∃ n. ((f ^ n))(∅)(x)

Rule argument

spec: monotone-set-fixpoint, name:subseteq-def, seq: ⊦ iset ⊆ iset0 ↔ (∀ x. iset(x) → iset0(x)), subst:[iset, iset0] → [f(λ x. ∃ n. ((f ^ n))(∅)(x)), λ x. ∃ n. ((f ^ n))(∅)(x)], <all paths>

Additional information

Used terms: f(λ x. ∃ n. ((f ^ n))(∅)(x)), λ x. ∃ n. ((f ^ n))(∅)(x)

Proof step 10 from proof iterate-is-fix (previous step) (back to tree)

⊦ iset ⊆ iset0 ↔ (∀ x. iset(x) → iset0(x))

A spec-lemma

Proof step 11 from proof iterate-is-fix (previous step) (back to tree)

cont(f), monotone(f) ⊦ ∀ x. f(λ x. ∃ n. ((f ^ n))(∅)(x))(x) → (∃ n. ((f ^ n))(∅)(x)), ∃ n. ((f ^ n))(∅)(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
cont(f), monotone(f), f(λ x. ∃ n. ((f ^ n))(∅)(x))(x0) ⊦ ∃ n. ((f ^ n))(∅)(x), ∃ n. ((f ^ n))(∅)(x0)

Rule argument

Additional information

Proof step 12 from proof iterate-is-fix (previous step) (back to tree)

cont(f), monotone(f), f(λ x. ∃ n. ((f ^ n))(∅)(x))(x0) ⊦ ∃ n. ((f ^ n))(∅)(x), ∃ n. ((f ^ n))(∅)(x0)

User applied rule apply lemma

And obtained the following new goals

Proof step 13:
cont(f) ⊦ (λ x0. ∃ n. ((f ^ n))(∅)(x0)) = f(λ x0. ∃ n. ((f ^ n))(∅)(x0))
Proof step 14:
cont(f), monotone(f), (λ x0. ∃ n. ((f ^ n))(∅)(x0)) = f(λ x0. ∃ n. ((f ^ n))(∅)(x0)), f(λ x. ∃ n. ((f ^ n))(∅)(x))(x0) ⊦ ∃ n. ((f ^ n))(∅)(x0), ∃ n. ((f ^ n))(∅)(x)

Rule argument

spec: <toplevel>, name:cont-iterate-fix, seq:cont(f) ⊦ (λ x0. ∃ n. ((f ^ n))(∅)(x0)) = f(λ x0. ∃ n. ((f ^ n))(∅)(x0)), subst:[f] → [f]

Additional information

Used terms: f

Proof step 13 from proof iterate-is-fix (previous step) (back to tree)

cont(f) ⊦ (λ x0. ∃ n. ((f ^ n))(∅)(x0)) = f(λ x0. ∃ n. ((f ^ n))(∅)(x0))

Lemma cont-iterate-fix

Proof step 14 from proof iterate-is-fix (previous step) (back to tree)

cont(f), monotone(f), (λ x0. ∃ n. ((f ^ n))(∅)(x0)) = f(λ x0. ∃ n. ((f ^ n))(∅)(x0)), f(λ x. ∃ n. ((f ^ n))(∅)(x))(x0) ⊦ ∃ n. ((f ^ n))(∅)(x0), ∃ n. ((f ^ n))(∅)(x)

User applied rule expand left

And closed the goal!

Rule argument

pos: left 3 terms: x0 abort: F discard: T computed: F

Additional information

Used terms: x0

Proof step 15 from proof iterate-is-fix (previous step) (back to tree)

cont(f), ¬ μ(f)(x), ∃ n. ((f ^ n))(∅)(x) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
((f ^ n))(∅)(x), cont(f), ¬ μ(f)(x) ⊦

Rule argument

Additional information

Proof step 16 from proof iterate-is-fix (previous step) (back to tree)

((f ^ n))(∅)(x), cont(f), ¬ μ(f)(x) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 17:
monotone(f) ⊦ ((f ^ n))(∅) ⊆ μ(f)
Proof step 18:
((f ^ n))(∅)(x), cont(f), ¬ monotone(f), ¬ μ(f)(x) ⊦
Proof step 20:
((f ^ n))(∅)(x), cont(f), ((f ^ n))(∅) ⊆ μ(f), ¬ μ(f)(x) ⊦

Rule argument

spec: <toplevel>, name:iterate-subseq-mu, seq:monotone(f) ⊦ ((f ^ n))(∅) ⊆ μ(f), subst:[f, n] → [f, n]

Additional information

Used terms: f, n

Proof step 17 from proof iterate-is-fix (previous step) (back to tree)

monotone(f) ⊦ ((f ^ n))(∅) ⊆ μ(f)

Lemma iterate-subseq-mu

Proof step 18 from proof iterate-is-fix (previous step) (back to tree)

((f ^ n))(∅)(x), cont(f), ¬ monotone(f), ¬ μ(f)(x) ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 19:
cont(f) ⊦ monotone(f)

Rule argument

spec: <toplevel>, name:cont-monotone, seq:cont(f) ⊦ monotone(f), subst:[f] → [f]

Additional information

Used terms: f

Proof step 19 from proof iterate-is-fix (previous step) (back to tree)

cont(f) ⊦ monotone(f)

Lemma cont-monotone

Proof step 20 from proof iterate-is-fix (previous step) (back to tree)

((f ^ n))(∅)(x), cont(f), ((f ^ n))(∅) ⊆ μ(f), ¬ μ(f)(x) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ iset ⊆ iset0 ↔ (∀ x. iset(x) → iset0(x))
Proof step 22:
((f ^ n))(∅)(x), cont(f), ¬ μ(f)(x), ∀ x. ((f ^ n))(∅)(x) → μ(f)(x) ⊦

Rule argument

spec: monotone-set-fixpoint, name:subseteq-def, seq: ⊦ iset ⊆ iset0 ↔ (∀ x. iset(x) → iset0(x)), subst:[iset, iset0] → [((f ^ n))(∅), μ(f)], <all paths>

Additional information

Used terms: ((f ^ n))(∅), μ(f)

Proof step 21 from proof iterate-is-fix (previous step) (back to tree)

⊦ iset ⊆ iset0 ↔ (∀ x. iset(x) → iset0(x))

A spec-lemma

Proof step 22 from proof iterate-is-fix (previous step) (back to tree)

((f ^ n))(∅)(x), cont(f), ¬ μ(f)(x), ∀ x. ((f ^ n))(∅)(x) → μ(f)(x) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: x

Additional information

Used terms: x

Proof Tree for iterate-is-fix (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.