Metamath Proof Explorer

Theorem tgcgrneq

Description: Congruence and equality. (Contributed by Thierry Arnoux, 27-Aug-2019)

Ref Expression
Hypotheses tkgeom.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
tkgeom.d ⊢ − = ( dist ‘ 𝐺 )
tkgeom.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
tkgeom.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
tgcgrcomlr.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
tgcgrcomlr.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
tgcgrcomlr.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
tgcgrcomlr.d ⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
tgcgrcomlr.6 ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐶 − 𝐷 ) )
tgcgrneq.1 ⊢ ( 𝜑 → 𝐴 ≠ 𝐵 )
Assertion tgcgrneq ( 𝜑 → 𝐶 ≠ 𝐷 )

Proof

Step Hyp Ref Expression
1 tkgeom.p ⊢ 𝑃 = ( Base ‘ 𝐺 )
2 tkgeom.d ⊢ − = ( dist ‘ 𝐺 )
3 tkgeom.i ⊢ 𝐼 = ( Itv ‘ 𝐺 )
4 tkgeom.g ⊢ ( 𝜑 → 𝐺 ∈ TarskiG )
5 tgcgrcomlr.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑃 )
6 tgcgrcomlr.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑃 )
7 tgcgrcomlr.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑃 )
8 tgcgrcomlr.d ⊢ ( 𝜑 → 𝐷 ∈ 𝑃 )
9 tgcgrcomlr.6 ⊢ ( 𝜑 → ( 𝐴 − 𝐵 ) = ( 𝐶 − 𝐷 ) )
10 tgcgrneq.1 ⊢ ( 𝜑 → 𝐴 ≠ 𝐵 )
11 1 2 3 4 5 6 7 8 9 tgcgreqb ⊢ ( 𝜑 → ( 𝐴 = 𝐵 ↔ 𝐶 = 𝐷 ) )
12 11 necon3bid ⊢ ( 𝜑 → ( 𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷 ) )
13 10 12 mpbid ⊢ ( 𝜑 → 𝐶 ≠ 𝐷 )