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 _0 'op_1' _1 ':' _1 '...' 'op_2' _2 ':' _2 ... '...' 'op_n' _n ':' _n Each `_i` is a proof for `_{i-1} op_i _i`. Recall that proofs are also expressions in Lean. The `_i` may also be of the form `{ }`, where `` is a proof for some equality `a = b`. The form `{ }` is just syntax sugar for Subst (Refl _{i-1}) That is, we are claiming we can obtain `_i` by replacing `a` with `b` in `_{i-1}`. Here is an example ```lean 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 `_i` do not need to be explicitly provided. We can use `by ` or `by ` 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. ```lean 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. ```lean 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. ```lean 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. ```