# mathlibdocumentation

tactic.monotonicity.interactive

meta inductive tactic.interactive.mono_function  :
(bool := tt) → Type

@[instance]

@[instance]

meta def tactic.interactive.ac_mono_ctx'.traverse {m : Type → Type} [applicative m] {α β : Type} :
(α → m β)

meta def tactic.interactive.ac_mono_ctx'.map {α β : Type} :

meta structure tactic.interactive.ac_mono_ctx'  :
Type → Type

@[instance]

(prefix,left,right,suffix) ← match_assoc unif l r finds the longest prefix and suffix common to l and r and returns them along with the differences

meta def tactic.interactive.bin_op  :
exprexpr

meta inductive tactic.interactive.mono_law  :
Type
• assoc :
• congr :
• other :

meta def tactic.interactive.mk_congr_args  :
exprexpr

meta def tactic.interactive.find_lemma  :
expr

def tactic.interactive.apply_rel {α : Sort u} (R : α → α → Sort v) {x y : α} (x' y' : α) :
R x yx = x'y = y'R x' y'

Equations
def tactic.interactive.list.minimum_on {α : Type u_1} {β : Type u_2} [linear_order β] :
(α → β)list αlist α

Equations
• (x :: xs) = (list.foldl (λ (_x : β × list α), tactic.interactive.list.minimum_on._match_1 f _x) (f x, [x]) xs).snd
• tactic.interactive.list.minimum_on._match_1 f (k, a) = λ (b : α), let k' : β := f b in ite (k < k') (k, a) (ite (k' < k) (k', [b]) (k, b :: a))
meta def tactic.interactive.best_match {β : Type} :
(exprtactic β)

• mono applies a monotonicity rule.
• mono* applies monotonicity rules repetitively.
• mono with x ≤ y or mono with [0 ≤ x,0 ≤ y] creates an assertion for the listed propositions. Those help to select the right monotonicity rule.
• mono left or mono right is useful when proving strict orderings: for x + y < w + z could be broken down into either
• left: x ≤ w and y < z or
• right: x < w and y ≤ z
• mono using [rule1,rule2] calls simp [rule1,rule2] before applying mono.
• The general syntax is mono '*'? ('with' hyp | 'with' [hyp1,hyp2])? ('using' [hyp1,hyp2])? mono_cfg?

To use it, first import tactic.monotonicity.

Here is an example of mono:

example (x y z k : ℤ)
(h : 3 ≤ (4 : ℤ))
(h' : z ≤ y) :
(k + 3 + x) - y ≤ (k + 4 + x) - z :=
begin
mono, -- unfold (-), apply add_le_add
{ -- ⊢ k + 3 + x ≤ k + 4 + x
-- ⊢ k + 3 ≤ k + 4
mono },
{ -- ⊢ -y ≤ -z
mono /- apply neg_le_neg -/ }
end


More succinctly, we can prove the same goal as:

example (x y z k : ℤ)
(h : 3 ≤ (4 : ℤ))
(h' : z ≤ y) :
(k + 3 + x) - y ≤ (k + 4 + x) - z :=
by mono*


transforms a goal of the form f x ≼ f y into x ≤ y using lemmas marked as monotonic.

Special care is taken when f is the repeated application of an associative operator and if the operator is commutative

(repeat_until_or_at_most n t u): repeat tactic t at most n times or until u succeeds

@[instance]

ac_mono reduces the f x ⊑ f y, for some relation ⊑ and a monotonic function f to x ≺ y.

ac_mono* unwraps monotonic functions until it can't.

ac_mono^k, for some literal number k applies monotonicity k times.

ac_mono h, with h a hypothesis, unwraps monotonic functions and uses h to solve the remaining goal. Can be combined with * or ^k: ac_mono* h

ac_mono : p asserts p and uses it to discharge the goal result unwrapping a series of monotonic functions. Can be combined with * or ^k: ac_mono* : p

In the case where f is an associative or commutative operator, ac_mono will consider any possible permutation of its arguments and use the one the minimizes the difference between the left-hand side and the right-hand side.

To use it, first import tactic.monotonicity.

ac_mono can be used as follows:

example (x y z k m n : ℕ)
(h₀ : z ≥ 0)
(h₁ : x ≤ y) :
(m + x + n) * z + k ≤ z * (y + n + m) + k :=
begin
ac_mono,
-- ⊢ (m + x + n) * z ≤ z * (y + n + m)
ac_mono,
-- ⊢ m + x + n ≤ y + n + m
ac_mono,
end


As with mono*, ac_mono* solves the goal in one go and so does ac_mono* h₁. The latter syntax becomes especially interesting in the following example:

example (x y z k m n : ℕ)
(h₀ : z ≥ 0)
(h₁ : m + x + n ≤ y + n + m) :
(m + x + n) * z + k ≤ z * (y + n + m) + k :=
by ac_mono* h₁.


By giving ac_mono the assumption h₁, we are asking ac_refl` to stop earlier than it would normally would.