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content/blog/quick-lean-if-then-else-in-hypothesis.md

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+---
+title: "Quick Lean: if-then-else statement in hypothesis"
+date: 2025-04-26T10:58:42-04:00
+draft: false
+tags: []
+math: false
+medium_enabled: false
+---
+
+When verifying proofs about code, I often end up with a hypothesis that has an if statement in it. Depending on how long it's been since I last used Lean, I forget how to deal with it. This is a quick post to remind me how.
+
+Example:
+
+```lean
+example (b: Bool) (n: Nat) (H1: if b then (n = 5) else (n = 3)) : n > 0 := by sorry
+```
+
+We want to show that for an arbitrary boolean and natural number `n`, that `n` will always be greater than zero. 
+
+Note that the condition of an arbitrary if statement can take a proposition instead of a boolean. However for this to work the proposition must be *decidable*. Checking the condition of a boolean is decidable, so we'll use that to simplify our example.
+
+From here, we use `by_cases (b = true)` to give us two new subgoals. One where it's true and one where it's false.
+
+```lean
+by_cases H2 : (b = true)
+case pos =>
+  sorry
+case neg =>
+  sorry
+```
+
+First let's consider the positive case. We can simplify `H1` to `n = 5` by using `H2` which states that `b` is true.
+
+```lean
+replace H1 : n = 5 := by
+  rwa [if_pos (by exact H2)] at H1
+```
+
+For this example, you don't need the parenthesis saying `(by exact H2)`. However, depending on your setup the proof that `b` is true might be too difficult for the rewrite system to infer. In those cases, you are required to specify the proof for it to work.
+
+Then, we can substitute `n` to have our subgoal as `5 > 0`.
+
+```lean
+subst n
+```
+
+This is the same as showing that `0 < 5`.
+
+```lean
+suffices 0 < 5 by exact gt_iff_lt.mpr this
+```
+
+From here we can apply one of the builtin theorems
+
+```lean
+exact Nat.zero_lt_succ 4
+```
+
+For the negative case, we will follow a similar pattern except instead of invoking `if_pos` to eliminate the if-statement, we will invoke `if_neg`.
+
+Thus, the complete proof for this is
+
+```lean
+example (b : Bool) (n : Nat) (H1: if b then (n = 5) else (n = 3)) : n > 0 := by
+  by_cases H2 : (b = true)
+  case pos =>
+    replace H1 : n = 5 := by
+      rwa [if_pos (by exact H2)] at H1
+    subst n
+    suffices 0 < 5 by exact gt_iff_lt.mpr this
+    exact Nat.zero_lt_succ 4
+  case neg =>
+    replace H1 : n = 3 := by
+      rwa [if_neg (by exact H2)] at H1
+    subst n
+    suffices 0 < 3 by exact gt_iff_lt.mpr this
+    exact Nat.zero_lt_succ 2
+```
+