LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Sign rules for products and monotonicity of multiplication

Statement

Let $F$ be an ordered field (Ordered field) with positive cone $P$ , and let $a, b, c \in F$ .

If $a > 0$ and $b > 0$ then $a b > 0$ .
If $a > 0$ and $b < 0$ then $a b < 0$ .
If $a < 0$ and $b < 0$ then $a b > 0$ .
If $c > 0$ then $a < b ⟺ a c < b c$ .
If $c < 0$ then $a < b ⟺ b c < a c$ (multiplication by a negative reverses order).

Statement references

Ordered field

Facts & Assumptions

Given: An ordered field $F$ with positive cone $P$ , and elements $a, b, c \in F$ .

[L1]

For $x, y \in F$ : $x < y ⟺ y - x \in P$ ; $x > 0 ⟺ x \in P$ ; $x < 0 ⟺ - x \in P$ ; and (trichotomy) for each $x$ exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds, so $0 \notin P$ and $x \in P \Rightarrow x \neq 0$ (Ordered field).

[L2]

$P$ is closed under multiplication: if $u, v \in P$ then $u v \in P$ (axiom O2 of Ordered field).

[L3]

In any field, $(- u) v = - (u v)$ and $(- u) (- v) = u v$ (Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ ).

[L4]

In any field, $0 \cdot u = 0$ (Multiplication by zero: $0 \cdot a = 0$ ).

Proof

technique · direct

1.1assume-hypL1L2

Assume $a > 0$ and $b > 0$ , i.e. $a \in P$ and $b \in P$ ; closure gives $a b \in P$ , so $a b > 0$ , proving claim 1.

1.2assume-hypL1L2

Assume $a > 0$ and $b < 0$ , i.e. $a \in P$ and $- b \in P$ ; closure gives $a (- b) \in P$ .

1.3assume-hypL1L2

Assume $a < 0$ and $b < 0$ , i.e. $- a \in P$ and $- b \in P$ ; closure gives $(- a) (- b) \in P$ .

1.4assume-hypL1L2

Assume $c > 0$ and $a < b$ , i.e. $c \in P$ and $b - a \in P$ ; closure gives $(b - a) c \in P$ .

1.5assume-hypL1L2

Assume $c < 0$ and $a < b$ , i.e. $- c \in P$ and $b - a \in P$ ; closure gives $(b - a) (- c) \in P$ .

2.1step 1.2L3L1

Since $a (- b) = - (a b)$ by L3, step 1.2 gives $- (a b) \in P$ , so $a b < 0$ , proving claim 2.

2.2step 1.3L3L1

Since $(- a) (- b) = a b$ by L3, step 1.3 gives $a b \in P$ , so $a b > 0$ , proving claim 3.

2.3step 1.4algebraL1

By distributivity $(b - a) c = b c - a c$ , so step 1.4 gives $b c - a c \in P$ , i.e. $a c < b c$ : this is the forward direction of claim 4.

2.4step 1.5L3algebraL1

By L3 and distributivity $(b - a) (- c) = - ((b - a) c) = - (b c - a c) = a c - b c$ , so step 1.5 gives $a c - b c \in P$ , i.e. $b c < a c$ : this is the forward direction of claim 5.

3.1assume-hypstep 2.3L4L1algebra

Conversely, assume $c > 0$ and $a c < b c$ ; by trichotomy exactly one of $a < b$ , $a = b$ , $b < a$ holds. If $a = b$ then $a c = b c$ , contradicting $a c < b c$ ; if $b < a$ then the argument of step 2.3 applied to the pair $b < a$ gives $b c < a c$ , so $b c - a c \in P$ (from $a c < b c$ ) while $a c - b c \in P$ (from $b c < a c$ ), putting the element $b c - a c$ and its negative $a c - b c$ both in $P$ , which trichotomy forbids; hence $a < b$ : this completes claim 4.

4.1assume-hypstep 2.4L4L1algebra∎

Conversely, assume $c < 0$ and $b c < a c$ ; by trichotomy exactly one of $a < b$ , $a = b$ , $b < a$ holds. If $a = b$ then $a c = b c$ , contradicting $b c < a c$ ; if $b < a$ then the argument of step 2.4 applied to the pair $b < a$ gives $a c < b c$ , so $a c - b c \in P$ (from $b c < a c$ ) while $b c - a c \in P$ (from $a c < b c$ ), putting the element $a c - b c$ and its negative $b c - a c$ both in $P$ , which trichotomy forbids; hence $a < b$ : this completes claim 5.

Sign rules for products and monotonicity of multiplication

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources