Reputation: 1632
I'm working my way through the chapter 4 of the lean tutorial.
I'd like to be able to prove simple equalities, such as a = b → a + 1 = b + 1
without having to use the calc environment. In other words I'd like to explicitly construct the proof term of:
example (a b : nat) (H1 : a = b) : a + 1 = b + 1 := sorry
My best guess is that I need to use eq.subst
and some relevant lemma about equality on natural numbers from the standard library, but I'm at loss. The closest lean example I can find is this:
example (A : Type) (a b : A) (P : A → Prop) (H1 : a = b) (H2 : P a) : P b :=
eq.subst H1 H2
Upvotes: 7
Views: 1527
Reputation: 15424
You can use the congrArg
lemma
lemma congrArg {α : Sort u} {β : Sort v} {a₁ a₂ : α} (f : α → β) :
a₁ = a₂ → f a₁ = f a₂
which means if you supply equal inputs to a function, the output values will be equal too.
The proof goes like this:
example (a b : Nat) (h : a = b) : a + 1 = b + 1 :=
congrArg (fun n => n + 1) h
Note, that Lean is able to infer that our function is fun n => n + 1
, so the proof can be simplified into congrArg _ h
.
Upvotes: 7
Reputation: 14734
Since you have an equality (a = b
), you can also rewrite the goal using tactic mode:
example (a b : nat) (H1 : a = b) : a + 1 = b + 1 :=
by rw H1
See Chapter 5 of Theorem Proving in Lean for an introduction to tactics.
Upvotes: 3
Reputation: 1347
More general : a = b -> a + c = b + c in a Ring
import algebra.ring
open algebra
variable {A : Type}
variables [s : ring A] (a b c : A)
include s
example (a b c : A) (H1 : a = b) : a + c = b + c :=
eq.subst H1 rfl
Upvotes: 1
Reputation: 1230
While congr_arg
is a good solution in general, this specific example can indeed be solved with eq.subst
+ higher-order unification (which congr_arg
uses internally).
example (a b : nat) (H1 : a = b) : a + 1 = b + 1 :=
eq.subst H1 rfl
Upvotes: 4