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Metamath Proof Explorer

Theorem bnj551

Description: First-order logic and set theory. (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Assertion bnj551 
|- ( ( m = suc p /\ m = suc i ) -> p = i )

Proof

Step Hyp Ref Expression
1 eqtr2 
 |-  ( ( m = suc p /\ m = suc i ) -> suc p = suc i )
2 suc11reg 
 |-  ( suc p = suc i <-> p = i )
3 1 2 sylib 
 |-  ( ( m = suc p /\ m = suc i ) -> p = i )