Theorem bnj551 34715

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

Assertion

Ref	Expression
bnj551	⊢ ((𝑚 = suc 𝑝 ∧ 𝑚 = suc 𝑖) → 𝑝 = 𝑖)

Proof of Theorem bnj551

Step	Hyp	Ref	Expression
1		eqtr2 2755	. 2 ⊢ ((𝑚 = suc 𝑝 ∧ 𝑚 = suc 𝑖) → suc 𝑝 = suc 𝑖)
2		suc11reg 9641	. 2 ⊢ (suc 𝑝 = suc 𝑖 ↔ 𝑝 = 𝑖)
3	1, 2	sylib 218	1 ⊢ ((𝑚 = suc 𝑝 ∧ 𝑚 = suc 𝑖) → 𝑝 = 𝑖)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1539  suc csuc 6365

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-12 2176  ax-ext 2706  ax-sep 5276  ax-nul 5286  ax-pr 5412  ax-un 7737  ax-reg 9614

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-ne 2932  df-ral 3051  df-rex 3060  df-rab 3420  df-v 3465  df-dif 3934  df-un 3936  df-in 3938  df-ss 3948  df-nul 4314  df-if 4506  df-pw 4582  df-sn 4607  df-pr 4609  df-op 4613  df-uni 4888  df-br 5124  df-opab 5186  df-eprel 5564  df-fr 5617  df-suc 6369

This theorem is referenced by:  bnj554  34872  bnj557  34874  bnj966  34917