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Theorem sbi1 2375
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 1867 . . . . 5 (𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑𝜑))
2 sbequ2 1867 . . . . 5 (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → (𝜑𝜓)))
31, 2syl5d 70 . . . 4 (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑𝜓)))
4 sbequ1 2093 . . . 4 (𝑥 = 𝑦 → (𝜓 → [𝑦 / 𝑥]𝜓))
53, 4syl6d 72 . . 3 (𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
65sps 2040 . 2 (∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
7 sb4 2339 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥]𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)))
8 sb4 2339 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ∀𝑥(𝑥 = 𝑦 → (𝜑𝜓))))
9 ax-2 7 . . . . . 6 ((𝑥 = 𝑦 → (𝜑𝜓)) → ((𝑥 = 𝑦𝜑) → (𝑥 = 𝑦𝜓)))
109al2imi 1731 . . . . 5 (∀𝑥(𝑥 = 𝑦 → (𝜑𝜓)) → (∀𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦𝜓)))
11 sb2 2335 . . . . 5 (∀𝑥(𝑥 = 𝑦𝜓) → [𝑦 / 𝑥]𝜓)
1210, 11syl6 34 . . . 4 (∀𝑥(𝑥 = 𝑦 → (𝜑𝜓)) → (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜓))
138, 12syl6 34 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → (∀𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜓)))
147, 13syl5d 70 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓)))
156, 14pm2.61i 174 1 ([𝑦 / 𝑥](𝜑𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wal 1472  [wsb 1865
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1711  ax-4 1726  ax-5 1825  ax-6 1873  ax-7 1920  ax-10 2004  ax-12 2031  ax-13 2228
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1866
This theorem is referenced by:  spsbim  2377  sbim  2378  2sb5ndVD  37967  2sb5ndALT  37989
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