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Theorem sbi1 2391
 Description: Removal of implication from substitution. (Contributed by NM, 14-May-1993.)
Assertion
Ref Expression
sbi1 ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))

Proof of Theorem sbi1
StepHypRef Expression
1 sbequ2 1879 . . . . 5 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
2 sbequ2 1879 . . . . 5 (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → (𝜑𝜓)))
31, 2syl5d 73 . . . 4 (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑𝜓)))
4 sbequ1 2107 . . . 4 (𝑥 = 𝑦 → (𝜓 → [𝑦 / 𝑥]𝜓))
53, 4syl6d 75 . . 3 (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
65sps 2053 . 2 (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
7 sb4 2355 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
8 sb4 2355 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑𝜓))))
9 ax-2 7 . . . . . 6 ((𝑥 = 𝑦 → (𝜑𝜓)) → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜓)))
109al2imi 1740 . . . . 5 (∀𝑥(𝑥 = 𝑦 → (𝜑𝜓)) → (∀𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜓)))
11 sb2 2351 . . . . 5 (∀𝑥(𝑥 = 𝑦𝜓) → [𝑦 / 𝑥]𝜓)
1210, 11syl6 35 . . . 4 (∀𝑥(𝑥 = 𝑦 → (𝜑𝜓)) → (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜓))
138, 12syl6 35 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜓)))
147, 13syl5d 73 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
156, 14pm2.61i 176 1 ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1478  [wsb 1877 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-10 2016  ax-12 2044  ax-13 2245 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1702  df-nf 1707  df-sb 1878 This theorem is referenced by:  spsbim  2393  sbim  2394  2sb5ndVD  38668  2sb5ndALT  38690
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