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PROOF 107

Prove.

Substitution and elimination give the same answer

On the real numbers ℝ.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

The solution set of a system is the set of all ordered pairs (x, y) that make both equations true.

Proof.

The solution set is determined by the system itself. It is independent of the method of solving.convention

∎

Definition D2

Two systems are equivalent when their solution sets are equal, and an operation turning one into the other is called an equivalent transformation.

Proof.

An equivalent transformation neither loses nor gains solutions.convention, D1

∎

Lemma L1

One step of elimination is an equivalent transformation.

Proof.

(1)One step of elimination turns (E₁, E₂) into (E₁, E₂ + kE₁).convention
(2)The two systems have the same solution set.(1), A1
(3)Hence it is an equivalent transformation.(2), D2

∎

Lemma L2

One step of substitution is an equivalent transformation.

Proof.

(1)Suppose a₁ ≠ 0. E₁ is equivalent to x = f(y), where f(y) is an expression in y alone.A3
(2)Substitution turns (x = f(y), E₂) into (x = f(y), E₂′), where E₂′ is E₂ with f(y) in place of x.(1), convention
(3)At a solution of this system, x = f(y) is true.(2), D1
(4)Hence replacing x by f(y) does not change the truth of E₂.(3), A2
(5)Conversely, replacing f(y) in E₂′ by x returns E₂. This is the same ground used in reverse.(3), A2
(6)The two systems have the same solution set.(4), (5), D1
(7)Hence it is an equivalent transformation.(6), D2

∎

Lemma L3

Chaining equivalent transformations gives an equivalent transformation again.

Proof.

(1)Suppose systems A and B are equivalent, and B and C are equivalent.assumption, D2
(2)The solution sets of A and of C are both equal to the solution set of B.(1), D2
(3)Hence the solution sets of A and C are equal.(2), A4

∎

Theorem T1

The system reached by elimination and the system reached by substitution have the same solution set.

Proof.

(1)Elimination passes only through equivalent transformations.L1
(2)Substitution also passes only through equivalent transformations.L2, A3
(3)The solution set of the system each method reaches equals the solution set of the original system.(1), (2), L3
(4)The solution set of the original system is determined uniquely by the system itself.D1
(5)The two results each equal the same single set, hence they equal each other.(3), (4), A4

∎

Theorem T2

If x = f(y) is discarded after substitution, the solution set grows.

Proof.

(1)Suppose (x = f(y), E₂′) has a solution, and let (x₀, y₀) be one.assumption, D1
(2)E₂′ contains no x.L2
(3)Hence for any real x, (x, y₀) also satisfies E₂′.(1), (2)
(4)f(y₀) + 1 differs from f(y₀); if they were equal, 1 = 0 would follow.A5, A6, A7
(5)Taking x = f(y₀) + 1, the pair (x, y₀) is a solution of E₂′ but does not satisfy x = f(y).(3), (4)
(6)Keeping only E₂′ enlarges the solution set. Substitution is an equivalent transformation only when x = f(y) is kept alongside.(5), D1, L2

∎

Remark

Can A2 be removed from this proof?

Find which line grounds the claim that substitution preserves the solution set. Separate the line that puts f(y) in from the line that takes it back out.