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Theorem spim 1699
Description: Specialization, using implicit substitution. Compare Lemma 14 of [Tarski] p. 70. The spim 1699 series of theorems requires that only one direction of the substitution hypothesis hold. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 3-Oct-2016.) (Proof rewritten by Jim Kingdon, 10-Jun-2018.)
Hypotheses
Ref Expression
spim.1 𝑥𝜓
spim.2 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
spim (∀𝑥𝜑𝜓)

Proof of Theorem spim
StepHypRef Expression
1 spim.1 . . 3 𝑥𝜓
21nfri 1482 . 2 (𝜓 → ∀𝑥𝜓)
3 spim.2 . 2 (𝑥 = 𝑦 → (𝜑𝜓))
42, 3spimh 1698 1 (∀𝑥𝜑𝜓)
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1312  wnf 1419
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1406  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-4 1470  ax-i9 1493  ax-ial 1497
This theorem depends on definitions:  df-bi 116  df-nf 1420
This theorem is referenced by:  cbv3  1703  chvar  1713  spimv  1765  2spim  12775
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