coq - How to apply rewrite inside a specific subexpression? -

i'm using online book "software foundations" learn coq.

in second chapter, asked prove "plus_assoc" theorem:

theorem plus_assoc : forall n m p : nat, n + (m + p) = (n + m) + p. 

i make use of 2 proven theorems:

theorem plus_comm : forall n m : nat, n + m = m + n. theorem plus_n_sm : forall n m : nat, s (n + m) = n + (s m). 

i prove plus_assoc theorem using induction on n:

proof.   intros n m p.   induction n [ | n' ].     reflexivity.      rewrite plus_comm.     rewrite <- plus_n_sm.     rewrite plus_comm.     rewrite ihn'.     symmetry.     rewrite plus_comm. 

at point, context (*) is:

1 subgoals case := "n = s n'" : string.string n' : nat m : nat p : nat ihn' : n' + (m + p) = n' + m + p ______________________________________(1/1) p + (s n' + m) = s (n' + m + p) 

i use plus_comm get

p + (m + s n') = s (n' + m + p) 

then plus_n_sm

p + s (m + n') = s (n' + m + p) 

then again plus_n_sm

s (p + (m + n')) = s (n' + m + p) 

and finish proof using plus_comm twice reflexivity

s (p + (n' + m)) = s (n' + m + p) s (n' + m + p) = s (n' + m + p) 

the small question don't know how apply plus_comm (s n' + m).

the great question is: why issuing

apply plus_comm. 

finishes proof instantly (in given context (*)) ?

thank in advance clarification !

fabian pijcke

you can apply plus_comm (s n' + m) instanciating (s n') , m.

    check plus_comm.     check plus_comm (s n').     check plus_comm (s n') m.     rewrite (plus_comm (s n') m).     rewrite <- plus_n_sm.     rewrite <- plus_n_sm.     rewrite (plus_comm m n').     rewrite plus_comm.     reflexivity. qed. 

i think using require import coq.setoids.setoid. , using rewrite plus_comm @ 2. supposed have same effect, doesn't work.

the reason apply plus_comm finishes goal because apply performs unification modulo conversion. say, p + (s n' + m) = s (n' + m + p) convertible p + (s n' + m) = s n' + m + p, , p + (s n' + m) = s n' + m + p unifiable ?1 + ?2 = ?2 + ?1.

in fact, if perform reduction using simpl tactic proof becomes shorter.

theorem plus_assoc : forall n m p : nat, n + (m + p) = (n + m) + p. proof. induction n.   reflexivity.    intros.   simpl.   apply f_equal.   apply ihn. qed.