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Theorem spimh 1667
Description: Specialization, using implicit substitition. Compare Lemma 14 of [Tarski] p. 70. The spim 1668 series of theorems requires that only one direction of the substitution hypothesis hold. (Contributed by NM, 5-Aug-1993.) (Revised by NM, 8-May-2008.) (New usage is discouraged.)
Hypotheses
Ref Expression
spimh.1 (𝜓 → ∀𝑥𝜓)
spimh.2 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
spimh (∀𝑥𝜑𝜓)

Proof of Theorem spimh
StepHypRef Expression
1 spimh.2 . . . 4 (𝑥 = 𝑦 → (𝜑𝜓))
2 spimh.1 . . . 4 (𝜓 → ∀𝑥𝜓)
31, 2syl6com 35 . . 3 (𝜑 → (𝑥 = 𝑦 → ∀𝑥𝜓))
43alimi 1385 . 2 (∀𝑥𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑥𝜓))
5 ax9o 1629 . 2 (∀𝑥(𝑥 = 𝑦 → ∀𝑥𝜓) → 𝜓)
64, 5syl 14 1 (∀𝑥𝜑𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1283
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-4 1441  ax-i9 1464  ax-ial 1468
This theorem depends on definitions:  df-bi 115
This theorem is referenced by:  spim  1668
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