Proof for mathd_algebra_513 - #47
Open
aleph-prover-test[bot] wants to merge 1 commit into
Open
Conversation
Automated commit at 20260127_180905
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Add this suggestion to a batch that can be applied as a single commit.This suggestion is invalid because no changes were made to the code.Suggestions cannot be applied while the pull request is closed.Suggestions cannot be applied while viewing a subset of changes.Only one suggestion per line can be applied in a batch.Add this suggestion to a batch that can be applied as a single commit.Applying suggestions on deleted lines is not supported.You must change the existing code in this line in order to create a valid suggestion.Outdated suggestions cannot be applied.This suggestion has been applied or marked resolved.Suggestions cannot be applied from pending reviews.Suggestions cannot be applied on multi-line comments.Suggestions cannot be applied while the pull request is queued to merge.Suggestion cannot be applied right now. Please check back later.
Proven lemmas: 1/1
The goal is to prove that if real numbers a and b satisfy the linear system 3a + 2b = 5 and a + b = 2, then necessarily a = 1 and b = 1. The proof is decomposed into two subgoals because the conclusion is a conjunction: first show a = 1, then show b = 1. The current proof completes both parts: it uses “constructor” to split the goal into the two equalities, and then applies linear arithmetic reasoning (linarith) to each, using the two given equations as input. Mathematically, this corresponds to solving the 2×2 system by elimination (for example, subtracting 2·(a + b = 2) from 3a + 2b = 5 to get a = 1, then substituting back to get b = 1). All subproblems are solved (2 out of 2), so nothing remains. An interesting aspect is that Lean’s linarith tactic can automatically carry out the elimination steps that one would do by hand for linear equations over ℝ.