Theorem tgextalt1

Description: tgexta 's FIRST conjunct, strict -- the remote interior angle at A is strictly less than the exterior angle at C . tgextale and tgextancgra deliver both halves already, but stated about <" C A B "> ; tgexta wants <" B A C "> , the same angle read backwards. So this is cgraswap plus two transports: leacgra1 for the leA half, and cgratr under mtand for the negation -- if <" B A C "> were congruent to the exterior angle then so would <" C A B "> be, which tgextancgra forbids. (Contributed by JAG, 12-Aug-2026)

	Ref	Expression
Hypotheses	tgextalt1.p	$$⊢P={\mathrm{Base}}_G$$
	tgextalt1.m	$$⊢\underset{\dot{}}{-}=\mathrm{dist}\left(G\right)$$
	tgextalt1.i	$$⊢I=\mathrm{Itv}\left(G\right)$$
	tgextalt1.l	$$⊢L={Line}_{𝒢}\left(G\right)$$
	tgextalt1.k	$$⊢K={hl}_{𝒢}\left(G\right)$$
	tgextalt1.s	$$⊢\underset{\dot{}}{\le }={\le }_{𝒢}^{\angle }\left(G\right)$$
	tgextalt1.g	$$⊢\phi \to G\in {𝒢}_{\mathrm{Tarski}}$$
	tgextalt1.a	$$⊢\phi \to A\in P$$
	tgextalt1.b	$$⊢\phi \to B\in P$$
	tgextalt1.c	$$⊢\phi \to C\in P$$
	tgextalt1.d	$$⊢\phi \to D\in P$$
	tgextalt1.ext	$$⊢\phi \to C\in \left(BID\right)$$
	tgextalt1.ncol	$$⊢\phi \to \neg \left(A\in \left(BLC\right)\vee B=C\right)$$
	tgextalt1.ne	$$⊢\phi \to C\ne D$$
Assertion	tgextalt1	$$⊢\phi \to \left(⟨“BAC”⟩\underset{\dot{}}{\le }⟨“ACD”⟩\wedge \neg ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)$$

Proof

Step	Hyp	Ref	Expression
1		tgextalt1.p	$$⊢P={\mathrm{Base}}_G$$
2		tgextalt1.m	$$⊢\underset{\dot{}}{-}=\mathrm{dist}\left(G\right)$$
3		tgextalt1.i	$$⊢I=\mathrm{Itv}\left(G\right)$$
4		tgextalt1.l	$$⊢L={Line}_{𝒢}\left(G\right)$$
5		tgextalt1.k	$$⊢K={hl}_{𝒢}\left(G\right)$$
6		tgextalt1.s	$$⊢\underset{\dot{}}{\le }={\le }_{𝒢}^{\angle }\left(G\right)$$
7		tgextalt1.g	$$⊢\phi \to G\in {𝒢}_{\mathrm{Tarski}}$$
8		tgextalt1.a	$$⊢\phi \to A\in P$$
9		tgextalt1.b	$$⊢\phi \to B\in P$$
10		tgextalt1.c	$$⊢\phi \to C\in P$$
11		tgextalt1.d	$$⊢\phi \to D\in P$$
12		tgextalt1.ext	$$⊢\phi \to C\in \left(BID\right)$$
13		tgextalt1.ncol	$$⊢\phi \to \neg \left(A\in \left(BLC\right)\vee B=C\right)$$
14		tgextalt1.ne	$$⊢\phi \to C\ne D$$
15	1 2 3 4 5 7 8 9 10 11 12 13 14	tgextale	$$⊢\phi \to ⟨“CAB”⟩{\le }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
16	1 3 4 7 8 9 10 13	ncolne2	$$⊢\phi \to A\ne C$$
17	16	necomd	$$⊢\phi \to C\ne A$$
18	1 3 4 7 8 9 10 13	ncolne1	$$⊢\phi \to A\ne B$$
19	1 3 7 5 10 8 9 17 18	cgraswap	$$⊢\phi \to ⟨“CAB”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“BAC”⟩$$
20	1 3 5 7 10 8 9 8 10 11 9 8 10 15 19	leacgra1	$$⊢\phi \to ⟨“BAC”⟩{\le }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
21	6	breqi	$$⊢⟨“BAC”⟩\underset{\dot{}}{\le }⟨“ACD”⟩\leftrightarrow ⟨“BAC”⟩{\le }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
22	20 21	sylibr	$$⊢\phi \to ⟨“BAC”⟩\underset{\dot{}}{\le }⟨“ACD”⟩$$
23	1 2 3 4 5 7 8 9 10 11 12 13 14	tgextancgra	$$⊢\phi \to \neg ⟨“CAB”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
24	7	adantr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to G\in {𝒢}_{\mathrm{Tarski}}$$
25	10	adantr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to C\in P$$
26	8	adantr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to A\in P$$
27	9	adantr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to B\in P$$
28	19	adantr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to ⟨“CAB”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“BAC”⟩$$
29	11	adantr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to D\in P$$
30		simpr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
31	1 3 24 5 25 26 27 27 26 25 28 26 25 29 30	cgratr	$$⊢\left(\phi \wedge ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)\to ⟨“CAB”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
32	23 31	mtand	$$⊢\phi \to \neg ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩$$
33	22 32	jca	$$⊢\phi \to \left(⟨“BAC”⟩\underset{\dot{}}{\le }⟨“ACD”⟩\wedge \neg ⟨“BAC”⟩{\sim }_{𝒢}^{\angle }\left(G\right)⟨“ACD”⟩\right)$$