PropositionStatement: 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.

Multiplying inequalities of positives

Statement

Let $F$ be an ordered field. If $0 < a < b$ and $0 < c < d$ then $a c < b d$ . If $0 \leq a \leq b$ and $0 \leq c \leq d$ then $a c \leq b d$ .

Facts & Assumptions

Given: Elements $a, b, c, d$ of an ordered field $F$ .

[L1]

For a positive multiplier $c > 0$ : $x < y ⟺ x c < y c$ ; multiplying a strict inequality by a positive element preserves it (Sign rules for products and monotonicity of multiplication).

[L2]

For $c \geq 0$ : $x \leq y \Rightarrow x c \leq y c$ ; the case $c > 0$ is [L1] and the case $c = 0$ gives $0 \leq 0$ (Sign rules for products and monotonicity of multiplication).

[L3]

The order is transitive: if $x < y$ and $y < z$ then $x < z$ , and likewise for $\leq$ , since the positive cone is closed under addition (Ordered field).

Proof

technique · direct

1.1L1

Assume $0 < a < b$ and $0 < c < d$ ; since $c > 0$ , scaling $a < b$ by $c$ gives $a c < b c$ .

1.2L1

Since $b > 0$ (from $0 < a < b$ ), scaling $c < d$ by $b$ gives $b c < b d$ .

1.3L2

Assume instead $0 \leq a \leq b$ and $0 \leq c \leq d$ ; since $c \geq 0$ , from $a \leq b$ we get $a c \leq b c$ .

1.4L2

Since $b \geq 0$ (from $0 \leq a \leq b$ ), from $c \leq d$ we get $b c \leq b d$ .

2.1L3step 1.1step 1.2

By transitivity, $a c < b c < b d$ , hence $a c < b d$ , proving the strict claim.

2.2L3step 1.3step 1.4

By transitivity of $\leq$ , $a c \leq b c \leq b d$ , hence $a c \leq b d$ , proving the nonstrict claim.

Multiplying inequalities of positives

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources