Theorem sbcco3g 2037

Description: Composition of two substitutions.

Hypothesis

Ref	Expression
sbcco3g.1	⊢ (x = A → B = C)

Assertion

Ref	Expression
sbcco3g	⊢ ((A ∈ R ⋀ ∀x B ∈ S) → ([A / x][B / y]φ ↔ [C / y]φ))

Distinct variable groups:   x,A   φ,x   x,C   x,y

Proof of Theorem sbcco3g

Step	Hyp	Ref	Expression
1		sbcnestg 2034	. 2 ⊢ ((A ∈ R ⋀ ∀x B ∈ S) → ([A / x][B / y]φ ↔ [[A / x]B / y]φ))
2		ax-17 969	. . . . . 6 ⊢ (z ∈ C → ∀x z ∈ C)
3	2	gen2 981	. . . . 5 ⊢ ∀x∀z(z ∈ C → ∀x z ∈ C)
4		sbcco3g.1	. . . . . 6 ⊢ (x = A → B = C)
5	4	ax-gen 961	. . . . 5 ⊢ ∀x(x = A → B = C)
6		csbiegft 2025	. . . . 5 ⊢ ((A ∈ R ⋀ ∀x∀z(z ∈ C → ∀x z ∈ C) ⋀ ∀x(x = A → B = C)) → [A / x]B = C)
7	3, 5, 6	mp3an23 906	. . . 4 ⊢ (A ∈ R → [A / x]B = C)
8		dfsbcq 1939	. . . 4 ⊢ ([A / x]B = C → ([[A / x]B / y]φ ↔ [C / y]φ))
9	7, 8	syl 10	. . 3 ⊢ (A ∈ R → ([[A / x]B / y]φ ↔ [C / y]φ))
10	9	adantr 389	. 2 ⊢ ((A ∈ R ⋀ ∀x B ∈ S) → ([[A / x]B / y]φ ↔ [C / y]φ))
11	1, 10	bitrd 527	1 ⊢ ((A ∈ R ⋀ ∀x B ∈ S) → ([A / x][B / y]φ ↔ [C / y]φ))

Syntax hints:   → wi 3   ↔ wb 146   ⋀ wa 223  ∀wal 952   = wceq 954   ∈ wcel 956  [wsbc 1168  [csb 1997

This theorem is referenced by:  fzshftralt 6462

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-9 963  ax-10 964  ax-11 965  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 776  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-v 1808  df-sbc 1938  df-csb 1998

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