Theorem sbal1 2062

Description: A theorem used in elimination of disjoint variable conditions on 𝑥, 𝑦 by replacing it with a distinctor ¬ ∀𝑥𝑥 = 𝑧. (Contributed by NM, 5-Aug-1993.) (Proof rewitten by Jim Kingdon, 24-Feb-2018.)

Assertion

Ref	Expression
sbal1	⊢ (¬ ∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)

Proof of Theorem sbal1

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbal 2060	. . . 4 ⊢ ([𝑤 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑤 / 𝑦]𝜑)
2	1	sbbii 1818	. . 3 ⊢ ([𝑧 / 𝑤][𝑤 / 𝑦]∀𝑥𝜑 ↔ [𝑧 / 𝑤]∀𝑥[𝑤 / 𝑦]𝜑)
3		sbal1yz 2061	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑤]∀𝑥[𝑤 / 𝑦]𝜑 ↔ ∀𝑥[𝑧 / 𝑤][𝑤 / 𝑦]𝜑))
4	2, 3	bitrid 192	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑤][𝑤 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑤][𝑤 / 𝑦]𝜑))
5		ax-17 1579	. . 3 ⊢ (∀𝑥𝜑 → ∀𝑤∀𝑥𝜑)
6	5	sbco2vh 2005	. 2 ⊢ ([𝑧 / 𝑤][𝑤 / 𝑦]∀𝑥𝜑 ↔ [𝑧 / 𝑦]∀𝑥𝜑)
7		ax-17 1579	. . . 4 ⊢ (𝜑 → ∀𝑤𝜑)
8	7	sbco2vh 2005	. . 3 ⊢ ([𝑧 / 𝑤][𝑤 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]𝜑)
9	8	albii 1523	. 2 ⊢ (∀𝑥[𝑧 / 𝑤][𝑤 / 𝑦]𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)
10	4, 6, 9	3bitr3g 222	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 105  ∀wal 1400  [wsb 1815

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588

This theorem depends on definitions:  df-bi 117  df-nf 1514  df-sb 1816

This theorem is referenced by: (None)