2.8 KiB
Calculational Proofs
A calculational proof is just a chain of intermediate results that are
meant to be composed by basic principles such as the transitivity of
=. In Lean, a calculation proof starts with the keyword calc, and has
the following syntax
calc <expr>_0 'op_1' <expr>_1 ':' <proof>_1
'...' 'op_2' <expr>_2 ':' <proof>_2
...
'...' 'op_n' <expr>_n ':' <proof>_n
Each <proof>_i is a proof for <expr>_{i-1} op_i <expr>_i.
Recall that proofs are also expressions in Lean. The <proof>_i
may also be of the form { <pr> }, where <pr> is a proof
for some equality a = b. The form { <pr> } is just syntax sugar
for
Subst (Refl <expr>_{i-1}) <pr>
That is, we are claiming we can obtain <expr>_i by replacing a with b
in <expr>_{i-1}.
Here is an example
Variables a b c d e : Nat.
Axiom Ax1 : a = b.
Axiom Ax2 : b = c + 1.
Axiom Ax3 : c = d.
Axiom Ax4 : e = 1 + d.
Theorem T : a = e
:= calc a = b : Ax1
... = c + 1 : Ax2
... = d + 1 : { Ax3 }
... = 1 + d : Nat::PlusComm d 1
... = e : Symm Ax4.
The proof expressions <proof>_i do not need to be explicitly provided.
We can use by <tactic> or by <solver> to (try to) automatically construct the
proof expression using the given tactic or solver.
Even when tactics and solvers are not used, we can still use the elaboration engine to fill
gaps in our calculational proofs. In the previous examples, we can use _ as arguments for the
Nat::PlusComm theorem. The Lean elaboration engine will infer d and 1 for us.
Here is the same example using placeholders.
Theorem T' : a = e
:= calc a = b : Ax1
... = c + 1 : Ax2
... = d + 1 : { Ax3 }
... = 1 + d : Nat::PlusComm _ _
... = e : Symm Ax4.
We can also use the operators =>, ⇒, <=>, ⇔ and ≠ in calculational proofs.
Here is an example.
Theorem T2 (a b c : Nat) (H1 : a = b) (H2 : b = c + 1) : a ≠ 0
:= calc a = b : H1
... = c + 1 : H2
... ≠ 0 : Nat::SuccNeZero _.
The Lean let construct can also be used to build calculational-like proofs.
Variable P : Nat → Nat → Bool.
Variable f : Nat → Nat.
Axiom Axf (a : Nat) : f (f a) = a.
Theorem T3 (a b : Nat) (H : P (f (f (f (f a)))) (f (f b))) : P a b
:= let s1 : P (f (f a)) (f (f b)) := Subst H (Axf a),
s2 : P a (f (f b)) := Subst s1 (Axf a),
s3 : P a b := Subst s2 (Axf b)
in s3.