[Imprint][Data Privacy Statement][E-Mail]



View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 1 from proof fun-neq (conclusion) (back to tree)

⊦ bf ≠ bf0 ↔ (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof fun-neq (previous step) (back to tree)

bf ≠ bf0 ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

User applied rule expand left

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Proof step 3 from proof fun-neq (previous step) (back to tree)

bf(n) ≠ bf0(n) ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

User applied rule induction

And obtained the following new goals

Rule argument

indvar: n wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: n

Proof step 4 from proof fun-neq (previous step) (back to tree)

bf(n) ≠ bf0(n), ∀ n1, bf, bf0. n1 < n → bf(n1) ≠ bf0(n1) → (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))) ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

User applied rule exists right

And obtained the following new goals

Rule argument

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

Additional information

Used terms: n

Proof step 5 from proof fun-neq (previous step) (back to tree)

bf(n) ≠ bf0(n), ∀ n1, bf, bf0. n1 < n → bf(n1) ≠ bf0(n1) → (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))) ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0)), bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 6 from proof fun-neq (previous step) (back to tree)

bf(n0) ≠ bf0(n0), bf(n) ≠ bf0(n), n0 < n, ∀ n1, bf, bf0. n1 < n → bf(n1) ≠ bf0(n1) → (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))) ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [n1, bf, bf0] → [n0, bf, bf0]

Additional information

Used terms: n0, bf, bf0

Proof step 7 from proof fun-neq (previous step) (back to tree)

bf(n0) ≠ bf0(n0), bf(n) ≠ bf0(n), n0 < n, ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0)), ∀ n1, bf, bf0. n1 < n → bf(n1) ≠ bf0(n1) → (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))) ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Proof step 8 from proof fun-neq (previous step) (back to tree)

bf(n0) ≠ bf0(n0), bf(n) ≠ bf0(n), bf(n1) ≠ bf0(n1), n0 < n, ∀ n0. n0 < n1 → bf(n0) = bf0(n0), ∀ n1, bf, bf0. n1 < n → bf(n1) ≠ bf0(n1) → (∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))) ⊦ ∃ n. bf(n) ≠ bf0(n) ∧ (∀ n0. n0 < n → bf(n0) = bf0(n0))

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: n1 abort: F discard: T computed: T

Additional information

Used terms: n1

Proof Tree for fun-neq (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