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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

cont(f) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule apply lemma

And obtained the following new goals

Rule argument

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

Additional information

Used terms: f

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

cont(f) ⊦ monotone(f)

Lemma cont-monotone

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:cont-def, seq:⊦ cont(f) ↔ (∀ chn. chain(chn) → (∀ x. f(λ x0. ∃ n. chn(n)(x0))(x) ↔ (∃ n. f(chn(n))(x)))), subst:[f] → [f], <all paths>

Additional information

Used terms: f

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

⊦ cont(f) ↔ (∀ chn. chain(chn) → (∀ x. f(λ x0. ∃ n. chn(n)(x0))(x) ↔ (∃ n. f(chn(n))(x))))

Lemma cont-def

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

monotone(f), ∀ chn. chain(chn) → (∀ x. f(λ x0. ∃ n. chn(n)(x0))(x) ↔ (∃ n. f(chn(n))(x))) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 2 terms: λ n. ((f ^ n))(λ x. false) abort: F discard: T computed: F

Additional information

Used terms: λ n. ((f ^ n))(λ x. false)

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

monotone(f), chain(λ n. ((f ^ n))(λ x. false)) → (∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x))) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

monotone(f), chain(λ n. ((f ^ n))(λ x. false)) → (∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x))) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

Additional information

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

monotone(f), ¬ chain(λ n. ((f ^ n))(λ x. false)) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:chain-def, seq: ⊦ chain(chn) ↔ (∀ m, n. chn(m) ⊆ chn(n) ∨ chn(n) ⊆ chn(m)), subst:[chn] → [λ n. ((f ^ n))(λ x. false)], <all paths>

Additional information

Used terms: λ n. ((f ^ n))(λ x. false)

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

⊦ chain(chn) ↔ (∀ m, n. chn(m) ⊆ chn(n) ∨ chn(n) ⊆ chn(m))

Lemma chain-def

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

monotone(f), ∃ m, n. ¬(((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false) ∨ ((f ^ n))(λ x. false) ⊆ ((f ^ m))(λ x. false)) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

monotone(f), ¬ ((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false), ¬ ((f ^ n))(λ x. false) ⊆ ((f ^ m))(λ x. false) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:iterate-subseq-leq, seq:monotone(f), m ≤ n ⊦ ((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false), subst:[f, m, n] → [f, m, n]

Additional information

Used terms: f, m, n

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

monotone(f), m ≤ n ⊦ ((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false)

Lemma iterate-subseq-leq

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

monotone(f), ¬ m ≤ n, ¬ ((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false), ¬ ((f ^ n))(λ x. false) ⊆ ((f ^ m))(λ x. false) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:iterate-subseq-leq, seq:monotone(f), m ≤ n ⊦ ((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false), subst:[f, m, n] → [f, n, m]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → ¬ n < m; (s) Used terms: f, n, m

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

monotone(f), m ≤ n ⊦ ((f ^ m))(λ x. false) ⊆ ((f ^ n))(λ x. false)

Lemma iterate-subseq-leq

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

monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 2 terms: x0 abort: F discard: F computed: T

Additional information

Used terms: x0

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

monotone(f), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ↔ (∃ n. f(((f ^ n))(λ x. false))(x0)), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ (∃ n. ((f ^ n))(λ x. false)(x0)) ↔ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

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

monotone(f), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ↔ (∃ n. f(((f ^ n))(λ x. false))(x0)), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)), ∃ n. ((f ^ n))(λ x. false)(x0) ⊦ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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

((f ^ n))(λ x. false)(x0), monotone(f), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ↔ (∃ n. f(((f ^ n))(λ x. false))(x0)), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 3

Additional information

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

((f ^ n))(λ x. false)(x0), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ ∃ n. f(((f ^ n))(λ x. false))(x0), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: n

Additional information

Used simplifier rules: iterate-base: ⊦ ((f ^ 0))(iset) = iset; nat freely generated by 0, +1;;

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

n = n0 +1, ((f ^ n))(λ x. false)(x0), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ ∃ n. f(((f ^ n))(λ x. false))(x0), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)

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

((f ^ (n0 + 1)))(λ x. false)(x0), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ ∃ n. f(((f ^ n))(λ x. false))(x0), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:iterate-rec, seq: ⊦ ((f ^ (n + 1)))(iset) = f(((f ^ n))(iset)), subst:[iset, n, f] → [λ x. false, n0, f], <all paths>

Additional information

Used terms: λ x. false, n0, f

Proof step 23 from proof cont-iterate-fix (previous step) (back to tree)

⊦ ((f ^ (n + 1)))(iset) = f(((f ^ n))(iset))

Lemma iterate-rec

Proof step 24 from proof cont-iterate-fix (previous step) (back to tree)

f(((f ^ n0))(λ x. false))(x0), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)) ⊦ ∃ n. f(((f ^ n))(λ x. false))(x0), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n0

Additional information

Used terms: n0

Proof step 25 from proof cont-iterate-fix (previous step) (back to tree)

monotone(f), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ↔ (∃ n. f(((f ^ n))(λ x. false))(x0)), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ⊦ ∃ n. ((f ^ n))(λ x. false)(x0)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

Additional information

Proof step 26 from proof cont-iterate-fix (previous step) (back to tree)

monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)), ∃ n. f(((f ^ n))(λ x. false))(x0), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ⊦ ∃ n. ((f ^ n))(λ x. false)(x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 27 from proof cont-iterate-fix (previous step) (back to tree)

f(((f ^ n))(λ x. false))(x0), monotone(f), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ⊦ ∃ n. ((f ^ n))(λ x. false)(x0)

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 1 terms: n + 1 abort: F discard: F computed: F

Additional information

Used terms: n + 1

Proof step 28 from proof cont-iterate-fix (previous step) (back to tree)

f(((f ^ n))(λ x. false))(x0), monotone(f), ¬ ((f ^ (n + 1)))(λ x. false)(x0), ∀ x. f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x) ↔ (∃ n. f(((f ^ n))(λ x. false))(x)), f(λ x0. ∃ n. ((f ^ n))(λ x. false)(x0))(x0) ⊦ ∃ n. ((f ^ n))(λ x. false)(x0)

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:iterate-rec, seq: ⊦ ((f ^ (n + 1)))(iset) = f(((f ^ n))(iset)), subst:[iset, n, f] → [λ x. false, n, f], <all paths>

Additional information

Used terms: λ x. false, n, f

Proof step 29 from proof cont-iterate-fix (previous step) (back to tree)

⊦ ((f ^ (n + 1)))(iset) = f(((f ^ n))(iset))

Lemma iterate-rec

Proof Tree for cont-iterate-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.

the proof tree