b181fd83ef
The `conv` tactic tries to close “trivial” goals after itself. As of now, it uses `try rfl`, which means it can close goals that are only trivial after reducing with default transparency. This is suboptimal * this can require a fair amount of unfolding, and possibly slow down the proof a lot. And the user cannot even prevent it. * it does not match what `rw` does, and a user might expect the two to behave the same. So this PR changes it to `with_reducible rfl`, matching `rw`’s behavior. I considered `with_reducible eq_refl` to only solve trivial goals that involve equality, but not other relations (e.g. `Perm xs xs`), but a discussion on mathlib pointed out that it’s expected and desirable to solve more general reflexive goals: https://leanprover.zulipchat.com/#narrow/stream/270676-lean4/topic/Closing.20after.20.60rw.60.2C.20.60conv.60.3A.20.60eq_refl.60.20instead.20of.20.60rfl.60/near/429851605
106 lines
2 KiB
Text
106 lines
2 KiB
Text
set_option pp.analyze false
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def p (x y : Nat) := x = y
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example (x y : Nat) : p (x + y) (y + x + 0) := by
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conv =>
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whnf
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congr
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. rfl
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. whnf; rfl
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trace_state
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rw [Nat.add_comm]
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rfl
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example (x y : Nat) : p (x + y) (y + x + 0) := by
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conv =>
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whnf
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rhs
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whnf
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trace_state
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rw [Nat.add_comm]
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rfl
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example (x y : Nat) : p (x + y) (y + x + 0) := by
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conv =>
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whnf
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lhs
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whnf
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conv =>
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rhs
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whnf
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trace_state
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apply Nat.add_comm x y
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example (x y : Nat) : p (x + y) (0 + y + x) := by
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conv =>
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whnf
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rhs
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rw [Nat.zero_add, Nat.add_comm]
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trace_state
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rfl
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done
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axiom div_self (x : Nat) : x ≠ 0 → x / x = 1
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example (h : x ≠ 0) : x / x + x = x.succ := by
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conv =>
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lhs
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arg 1
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rw [div_self]
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rfl
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tactic => assumption
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done
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show 1 + x = x.succ
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rw [Nat.succ_add, Nat.zero_add]
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example (h1 : x ≠ 0) (h2 : y = x / x) : y = 1 := by
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conv at h2 =>
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rhs
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rw [div_self]
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rfl
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tactic => assumption
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assumption
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example : id (fun x => 0 + x) = id := by
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conv =>
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lhs
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arg 1
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ext y
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rw [Nat.zero_add]
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rfl
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def f (x : Nat) :=
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if x > 0 then
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x + 1
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else
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x + 2
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example (g : Nat → Nat) (h₁ : g x = x + 1) (h₂ : x > 0) : g x = f x := by
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conv =>
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rhs
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simp [f, h₂]
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exact h₁
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example (h₁ : f x = x + 1) (h₂ : x > 0) : f x = f x := by
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conv =>
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rhs
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simp [f, h₂]
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exact h₁
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example (x y : Nat) (f : Nat → Nat → Nat) (g : Nat → Nat) (h₁ : ∀ z, f z z = z) (h₂ : ∀ x y, f (g x) (g y) = y) : f (g (0 + y)) (f (g x) (g (x + 0))) = x := by
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conv in _ + 0 => apply Nat.add_zero
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trace_state
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conv in 0 + _ => apply Nat.zero_add
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trace_state
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simp [h₁, h₂]
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example (x y : Nat) (f : Nat → Nat → Nat) (g : Nat → Nat)
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(h₁ : ∀ z, f z z = z) (h₂ : ∀ x y, f (g x) (g y) = y)
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(h₃ : f (g (0 + x)) (g x) = 0)
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: g x = 0 := by
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conv at h₃ in 0 + x => apply Nat.zero_add
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trace_state
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conv at h₃ => lhs; apply h₁
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trace_state
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assumption
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