Section strictsub_trans1a.
Hypothesis m : EQ A C.
Let step2 : SUB C A. Proof EQ_backwards A C m.
Let step3 : SUB B A. Proof SUB_trans B C A k step2.
Let step4 : EQ A B. Proof Build_EQ A B h1 step3.

Lemma strictsub_trans1a_Out : False.