fix: suggest (rfl) not id rfl in linter (#13319)

This PR amends #13317 to suggest `:= (rfl)` as the recommended way to
avoid a theorem to be automatically marked `[defeq]`, for consistency
with existing documentation. Rationale: the special treatment of `:=
rfl` is based on syntax, not the proof term, so it’s appropriate to use
different syntax. And also I like the way it reads like a “muted whisper
of `rfl`”.

src/Init/Data/Nat/Gcd.lean

@@ -60,7 +60,7 @@ theorem gcd_def (x y : Nat) : gcd x y = if x = 0 then y else gcd (y % x) x := by
   cases n with
   | zero => simp
   | succ n =>
-    -- `simp [gcd_succ]` produces an invalid term unless `gcd_succ` is proved with `id rfl` instead
+    -- `simp [gcd_succ]` produces an invalid term unless `gcd_succ` is proved with `(rfl)` instead
     rw [gcd_succ]
     exact gcd_zero_left _
 instance : Std.LawfulIdentity gcd 0 where

src/Lean/Meta/Tactic/Simp/SimpTheorems.lean

@@ -382,10 +382,10 @@ private def mkSimpTheoremCore (origin : Origin) (e : Expr) (levelParams : Array
     forallTelescopeReducing type fun _ type => do
       let checkDefEq (lhs rhs : Expr) := do
         unless (← withTransparency .instances <| isDefEq lhs rhs) do
-          logWarning m!"`{origin.key}` is a `rfl` simp theorem, but its left-hand side{indentExpr lhs}\n\
+          logWarning m!"`{origin.key}` is a `[defeq]` simp theorem, but its left-hand side{indentExpr lhs}\n\
             is not definitionally equal to the right-hand side{indentExpr rhs}\n\
             at `.instances` transparency. Possible solutions:\n\
-            1- use `id rfl` as the proof\n\
+            1- use `(rfl)` as the proof\n\
             2- mark constants occurring in the lhs and rhs as `[implicit_reducible]`"
       match_expr type with
       | Eq _ lhs rhs => checkDefEq lhs rhs

tests/elab/simp_rfl_check_transparency.lean

@@ -5,19 +5,19 @@ def myId (n : Nat) : Nat := n
 
 -- LHS `myId n` and RHS `n` are only defeq if `myId` is unfolded (requires default transparency)
 /--
-warning: `myId_eq` is a `rfl` simp theorem, but its left-hand side
+warning: `myId_eq` is a `[defeq]` simp theorem, but its left-hand side
   myId n
 is not definitionally equal to the right-hand side
   n
 at `.instances` transparency. Possible solutions:
-1- use `id rfl` as the proof
+1- use `(rfl)` as the proof
 2- mark constants occurring in the lhs and rhs as `[implicit_reducible]`
 -/
 #guard_msgs in
 @[simp] theorem myId_eq (n : Nat) : myId n = n := rfl
 
 #guard_msgs in
-@[simp] theorem myId_eq' (n : Nat) : myId n = n := id rfl
+@[simp] theorem myId_eq' (n : Nat) : myId n = n := (rfl)
 
 attribute [implicit_reducible] myId
 
@@ -32,16 +32,16 @@ attribute [implicit_reducible] myId
 @[simp] theorem myId2_eq (n : Nat) : myId2 n = n := rfl
 
 /--
-warning: `add_zero` is a `rfl` simp theorem, but its left-hand side
+warning: `add_zero` is a `[defeq]` simp theorem, but its left-hand side
   n + 0
 is not definitionally equal to the right-hand side
   n
 at `.instances` transparency. Possible solutions:
-1- use `id rfl` as the proof
+1- use `(rfl)` as the proof
 2- mark constants occurring in the lhs and rhs as `[implicit_reducible]`
 -/
 #guard_msgs in
 @[simp] theorem add_zero (n : Nat) : n + 0 = n := rfl
 
 #guard_msgs in
-@[simp] theorem add_zero' (n : Nat) : n + 0 = n := id rfl
+@[simp] theorem add_zero' (n : Nat) : n + 0 = n := (rfl)