lt_sub_left_input_iff_of_balanced
plain-language theorem explainer
If two signed orbits represent the same integer, subtracting a fixed third orbit from either yields differences that stand in the same strict order relative to any fourth orbit. Order theorists building the PRC integer surface cite it when transporting inequalities across balanced representatives. The proof is a one-line composition of left-subtraction congruence with left-order congruence under balance.
Claim. Let $a,a',b,c$ be signed orbits. If $a$ and $a'$ are balanced (i.e. $a_{\mathrm{pos}}+a'_{\mathrm{neg}}=a'_{\mathrm{pos}}+a_{\mathrm{neg}}$), then $a-b < c$ if and only if $a'-b < c$, where $<$ is the internal strict order (nonnegative nonzero difference).
background
In the Primitive Recognition Calculus, integers are realized as signed orbits: pairs of nonnegative orbit positions recording positive and negative legs. Two signed orbits are balanced when $a.pos + b.neg = b.pos + a.neg$; this is the internal equivalence that identifies representatives of the same integer, defined entirely on $\delta$-orbit data.
Strict order on signed orbits is the internal relation "nonnegative difference and not balanced": $a < b$ means $a \le b$ and $\neg,\mathrm{balanced}(a,b)$. Subtraction is the corresponding signed-orbit operation. The local module develops the order surface on these objects so that the integer structure forced from recognition logic is closed under the usual comparison and arithmetic laws.
Upstream, left-order congruence under balance states that balanced $a\sim a'$ implies $a < b \leftrightarrow a' < b$. Left-subtraction congruence states that balanced $a\sim a'$ implies $\mathrm{sub}(a,b)\sim\mathrm{sub}(a',b)$. Both reduce through the toInt embedding when available.
proof idea
Term-mode one-liner. First apply left-subtraction congruence under balance to the hypothesis $a\sim a'$, obtaining $\mathrm{sub}(a,b)\sim\mathrm{sub}(a',b)$. Feed that balanced pair into left-order congruence under balance with right operand $c$. The result is exactly $\mathrm{sub}(a,b) < c \leftrightarrow \mathrm{sub}(a',b) < c$. No further rewriting or case analysis.
why it matters
Feeds the integer-order certificate, whose doc-comment records that "the internal signed-orbit order surface is closed." That certificate packages display lemmas (truncated subtraction, $\le$ flags, absolute difference, nonnegative flags) needed to expose the PRC integer order to downstream arithmetic and forcing layers.
Within Recognition Science foundation work, signed-orbit order is the bridge from pure recognition logic (LogicNat strict order via successor witnesses) to the integer rung of the forcing chain. Congruence of inequalities under balanced left inputs is a routine but mandatory closure fact: without it, order statements would depend on the choice of representative rather than the integer class. It does not itself invoke T5–T8 landmarks; it sits earlier, sealing the integer substrate those later steps presuppose.
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