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Theorem ax11i 1602
Description: Inference that has ax-11 1397 (without 𝑦) as its conclusion and doesn't require ax-10 1396, ax-11 1397, or ax-12 1402 for its proof. The hypotheses may be eliminable without one or more of these axioms in special cases. Proof similar to Lemma 16 of [Tarski] p. 70. (Contributed by NM, 20-May-2008.)
Hypotheses
Ref Expression
ax11i.1 (𝑥 = 𝑦 → (𝜑𝜓))
ax11i.2 (𝜓 → ∀𝑥𝜓)
Assertion
Ref Expression
ax11i (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))

Proof of Theorem ax11i
StepHypRef Expression
1 ax11i.1 . 2 (𝑥 = 𝑦 → (𝜑𝜓))
2 ax11i.2 . . 3 (𝜓 → ∀𝑥𝜓)
31biimprcd 149 . . 3 (𝜓 → (𝑥 = 𝑦𝜑))
42, 3alrimih 1358 . 2 (𝜓 → ∀𝑥(𝑥 = 𝑦𝜑))
51, 4syl6bi 152 1 (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98  wal 1241
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338
This theorem depends on definitions:  df-bi 110
This theorem is referenced by: (None)
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