lean4-htt/tests/lean/run/substlet.lean
jrr6 34bd6e8bfd
feat: improve split error messages (#9424)
This PR improves the error messages produced by the `split` tactic,
including suggesting syntax fixes and related tactics with which it
might be confused.

Note that, to avoid clashing with the new error message styling
conventions used in these messages, this PR also updates the formatting
of the message produced by `throwTacticEx`.

Closes #6224
2025-07-18 22:36:10 +00:00

113 lines
1.8 KiB
Text

/-!
# Tests of the `subst` tactic when `let`s are present.
-/
/-!
Eliminates `a` even though `e : id a = m`.
-/
/--
trace: case intro
n : Nat
m : Nat := n
a : Nat
e : id a = m
⊢ 0 + n = n
---
trace: case intro
a : Nat
m : Nat := id a
⊢ 0 + id a = id a
-/
#guard_msgs in
theorem ex1 (n : Nat) : 0 + n = n := by
let m := n
have h : ∃ k, id k = m := ⟨m, rfl⟩
cases h with
| intro a e =>
trace_state
subst e
trace_state
apply Nat.zero_add
/-!
Eliminates `a` even though `e : m = id a`.
-/
/--
trace: case intro
n : Nat
m : Nat := n
a : Nat
e : m = id a
⊢ 0 + n = n
---
trace: case intro
n : Nat
m : Nat := n
⊢ 0 + n = n
-/
#guard_msgs in
theorem ex2 (n : Nat) : 0 + n = n := by
let m := n
have h : ∃ k, m = id k := ⟨m, rfl⟩
cases h with
| intro a e =>
trace_state
subst e
trace_state
apply Nat.zero_add
/-!
Since `v` is a let binding, the `subst v` tactic instead
zeta delta reduces it everywhere and then clears it.
-/
/--
trace: n : Nat
h : n = 0
m : Nat := n + 1
v : Nat := m + 1
this : v = n + 2
⊢ 0 + n = 0
---
trace: n : Nat
h : n = 0
m : Nat := n + 1
this : m + 1 = n + 2
⊢ 0 + n = 0
---
trace: m : Nat := 0 + 1
this : m + 1 = 0 + 2
⊢ 0 + 0 = 0
-/
#guard_msgs in
theorem ex3 (n : Nat) (h : n = 0) : 0 + n = 0 := by
let m := n + 1
let v := m + 1
have : v = n + 2 := rfl
trace_state
subst v
trace_state
subst n
trace_state
rfl
/-!
Can't do `subst this` with `this : v = n + 2` since `v` is a let binding.
The tactic sees `m + 1 = n + 2` and fails.
-/
/--
error: Tactic `subst` failed: invalid equality proof, it is not of the form (x = t) or (t = x)
v = n + 2
n : Nat
h : n = 0
m : Nat := n + 1
v : Nat := m + 1
this : v = n + 2
⊢ 0 + n = 0
-/
#guard_msgs in
theorem ex4 (n : Nat) (h : n = 0) : 0 + n = 0 := by
let m := n + 1
let v := m + 1
have : v = n + 2 := rfl
subst this