Theorem spsbim 1721

Description: Specialization of implication. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 21-Jan-2018.)

Assertion

Ref	Expression
spsbim	⊢ (∀x(φ → ψ) → ([y / x]φ → [y / x]ψ))

Proof of Theorem spsbim

Step	Hyp	Ref	Expression
1		imim2 49	. . . 4 ⊢ ((φ → ψ) → ((x = y → φ) → (x = y → ψ)))
2	1	sps 1427	. . 3 ⊢ (∀x(φ → ψ) → ((x = y → φ) → (x = y → ψ)))
3		id 19	. . . . . 6 ⊢ ((φ → ψ) → (φ → ψ))
4	3	anim2d 320	. . . . 5 ⊢ ((φ → ψ) → ((x = y ∧ φ) → (x = y ∧ ψ)))
5	4	alimi 1341	. . . 4 ⊢ (∀x(φ → ψ) → ∀x((x = y ∧ φ) → (x = y ∧ ψ)))
6		exim 1487	. . . 4 ⊢ (∀x((x = y ∧ φ) → (x = y ∧ ψ)) → (∃x(x = y ∧ φ) → ∃x(x = y ∧ ψ)))
7	5, 6	syl 14	. . 3 ⊢ (∀x(φ → ψ) → (∃x(x = y ∧ φ) → ∃x(x = y ∧ ψ)))
8	2, 7	anim12d 318	. 2 ⊢ (∀x(φ → ψ) → (((x = y → φ) ∧ ∃x(x = y ∧ φ)) → ((x = y → ψ) ∧ ∃x(x = y ∧ ψ))))
9		df-sb 1643	. 2 ⊢ ([y / x]φ ↔ ((x = y → φ) ∧ ∃x(x = y ∧ φ)))
10		df-sb 1643	. 2 ⊢ ([y / x]ψ ↔ ((x = y → ψ) ∧ ∃x(x = y ∧ ψ)))
11	8, 9, 10	3imtr4g 194	1 ⊢ (∀x(φ → ψ) → ([y / x]φ → [y / x]ψ))

Syntax hints:   → wi 4   ∧ wa 97  ∀wal 1240  ∃wex 1378  [wsb 1642

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 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-4 1397  ax-ial 1424

This theorem depends on definitions:  df-bi 110  df-sb 1643

This theorem is referenced by:  spsbbi  1722  hbsb4t  1886  moim  1961