Theorem legeq

Description: Deduce equality from "less than" null segments. (Contributed by Thierry Arnoux, 12-Aug-2019)

	Ref	Expression
Hypotheses	legval.p	⊢ 𝑃 = ( Base ‘ 𝐺 )
	legval.d	⊢ − = ( dist ‘ 𝐺 )
	legval.i	⊢ 𝐼 = ( Itv ‘ 𝐺 )
	legval.l	⊢ ≤ = ( ≤G ‘ 𝐺 )
	legval.g	⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
	legid.a	⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
	legid.b	⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
	legtrd.c	⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
	legtrd.d	⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
	legeq.1	⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) ≤ ( 𝐶 − 𝐶 ) )
Assertion	legeq	⊢ ( 𝜑 → 𝐴 = 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		legval.p	⊢ 𝑃 = ( Base ‘ 𝐺 )
2		legval.d	⊢ − = ( dist ‘ 𝐺 )
3		legval.i	⊢ 𝐼 = ( Itv ‘ 𝐺 )
4		legval.l	⊢ ≤ = ( ≤G ‘ 𝐺 )
5		legval.g	⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
6		legid.a	⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
7		legid.b	⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
8		legtrd.c	⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
9		legtrd.d	⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
10		legeq.1	⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) ≤ ( 𝐶 − 𝐶 ) )
11	1 2 3 4 5 8 6 6 7	leg0	⊢ ( 𝜑 → ( 𝐶 − 𝐶 ) ≤ ( 𝐴 − 𝐵 ) )
12	1 2 3 4 5 6 7 8 8 10 11	legtri3	⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐶 − 𝐶 ) )
13	1 2 3 5 6 7 8 12	axtgcgrid	⊢ ( 𝜑 → 𝐴 = 𝐵 )