Arithmetic expressions and big-step semantic

Question

I am working on a project where i have to define a type for arithmetic expressions and an inductive predicate for big-step operational semantic.

First i began with defining a type for valuations and i decided to do a list of string * Z

Definition valuation  := list (string * Z).

then i defined a function which is used to apply the valuation

Fixpoint apply_valuation (s : string) (v : valuation) : Z := 
match v with 
| nil => 0
| cons (s',v') q => if  (string_dec s' s) then v'
                else (apply_valuation s q)
end.

and this is how i defined my arithmetic expressions type :

Inductive EA : Type := 
| Cst (n : Z)
| Var (v : string) 
| Plus (a1 a2 : EA) 
| Minus (a1 a2 : EA) 
| Mult (a1 a2 : EA)
| Div (a1 a2 : EA).

And my inductive predicate for big-step semantic is

 Inductive Sem_AE (s : valuation) : EA -> Z -> Prop :=
| cst_sem : forall z, Sem_AE s (Cst z) z
| var_sem : forall v , Sem_AE s (Var v) (apply_valuation v s)
| plus_sem : forall e1 e2 z1 z2 , 
          Sem_AE s e1 z1 -> Sem_AE s e2 z2 -> Sem_AE s (Plus e1 e2) (z1 + z2)
| minus_sem : forall e1 e2 z1 z2 , 
         Sem_AE s e1 z1 ->  Sem_AE s e2 z2 ->  Sem_AE s (Minus e1 e2) (z1 - z2)
| mult_sem: forall e1 e2 z1 z2 , 
         Sem_AE s e1 z1 ->   Sem_AE s e2 z2 ->  Sem_AE s (Mult e1 e2) (z1 * z2)
| sem_Div : forall e1 e2 z1 z2 , 
         Sem_AE s e1 z1 ->  Sem_AE s e2 z2 -> Sem_AE s (Div e1 e2) (z1 / z2).

Now i have to prove that big-step semantic is deteministic, but i get stuck when it arrives to prove it for plus

Lemma sem_det: forall s a z z', Sem_AE s a z -> Sem_AE s a z' -> z = z'.
Proof.
intros. induction H.
- inversion H0. reflexivity.
- inversion H0. reflexivity.
- inversion H0. subst.

This is what i have in my context

1 subgoals
s : valuation
e1 : EA
e2 : EA
z1 : Z
z2 : Z
H : Sem_EA s e1 z1
H1 : Sem_EA s e2 z2
z0 : Z
z3 : Z
H4 : Sem_EA s e1 z0
H6 : Sem_EA s e2 z3
H0 : Sem_EA s (Plus e1 e2) (z0 + z3)
IHSem_EA1 : Sem_EA s e1 (z0 + z3) -> z1 = z0 + z3
IHSem_EA2 : Sem_EA s e2 (z0 + z3) -> z2 = z0 + z3
______________________________________(1/1)
 z1 + z2 = z0 + z3

Any ideas how to continue this proof? and am i in the right path?

Thanks a lot.

score 3 · Accepted Answer · answered Dec 04 '20 at 14:30

This is a common gotcha. The problem is that your induction hypothesis is not general enough. You can fix this issue by generalizing over z':

Lemma sem_det: forall s a z z', Sem_AE s a z -> Sem_AE s a z' -> z = z'.
Proof.
intros. generalize dependent z'. induction H; intros.
- inversion H0. reflexivity.
- inversion H0. reflexivity.
- inversion H1. subst.
  rewrite (IHSem_AE1 _ H4).
  now rewrite (IHSem_AE2 _ H6).

It is instructive to compare the state of the generalized proof against your previous attempt. You can read the Software Foundations book for more information.

Small tip: your proofs are relying on the names that Coq generates for you. You should never do this, since these choices are brittle and cause your proof scripts to break easily.

Arithmetic expressions and big-step semantic

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