* 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 #+BEGIN_SRC calc <expr>_0 'op_1' <expr>_1 ':' <proof>_1 '...' 'op_2' <expr>_2 ':' <proof>_2 ... '...' 'op_n' <expr>_n ':' <proof>_n #+END_SRC 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 =eq.subst <pr> (refl <expr>_{i-1})= That is, we are claiming we can obtain =<expr>_i= by replacing =a= with =b= in =<expr>_{i-1}=. Here is an example #+BEGIN_SRC lean import data.nat open nat constants 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 : add.comm d 1 ... = e : eq.symm Ax4. #+END_SRC 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. The Lean elaboration engine infers them automatically for us. The =calc= command can be configured for any relation that supports some form of transitivity. It can even combine different relations. #+BEGIN_SRC lean import data.nat open nat theorem T2 (a b c : nat) (H1 : a = b) (H2 : b = c + 1) : a ≠ 0 := calc a = b : H1 ... = c + 1 : H2 ... = succ c : add.one c ... ≠ 0 : succ_ne_zero c #+END_SRC The Lean simplifier can be used to reduce the size of calculational proofs. In the following example, we create a rewrite rule set with basic theorems from the Natural number library, and some of the axioms declared above.