Metamath Proof Explorer

Theorem 2sb6rf

Description: Reversed double substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 3-Feb-2005) (Revised by Mario Carneiro, 6-Oct-2016) Remove variable constraints. (Revised by Wolf Lammen, 28-Sep-2018) (Proof shortened by Wolf Lammen, 13-Apr-2023) (New usage is discouraged.)

Ref Expression
Hypotheses 2sb5rf.1 ⊢ Ⅎ z φ
2sb5rf.2 ⊢ Ⅎ w φ
Assertion 2sb6rf ⊢ φ ↔ ∀ z ∀ w z = x ∧ w = y → z x w y φ

Proof

Step Hyp Ref Expression
1 2sb5rf.1 ⊢ Ⅎ z φ
2 2sb5rf.2 ⊢ Ⅎ w φ
3 1 19.23 ⊢ ∀ z ∃ w z = x ∧ w = y → φ ↔ ∃ z ∃ w z = x ∧ w = y → φ
4 2 19.23 ⊢ ∀ w z = x ∧ w = y → φ ↔ ∃ w z = x ∧ w = y → φ
5 4 albii ⊢ ∀ z ∀ w z = x ∧ w = y → φ ↔ ∀ z ∃ w z = x ∧ w = y → φ
6 2ax6e ⊢ ∃ z ∃ w z = x ∧ w = y
7 6 a1bi ⊢ φ ↔ ∃ z ∃ w z = x ∧ w = y → φ
8 3 5 7 3bitr4ri ⊢ φ ↔ ∀ z ∀ w z = x ∧ w = y → φ
9 sbequ12r ⊢ z = x → z x w y φ ↔ w y φ
10 sbequ12r ⊢ w = y → w y φ ↔ φ
11 9 10 sylan9bb ⊢ z = x ∧ w = y → z x w y φ ↔ φ
12 11 pm5.74i ⊢ z = x ∧ w = y → z x w y φ ↔ z = x ∧ w = y → φ
13 12 2albii ⊢ ∀ z ∀ w z = x ∧ w = y → z x w y φ ↔ ∀ z ∀ w z = x ∧ w = y → φ
14 8 13 bitr4i ⊢ φ ↔ ∀ z ∀ w z = x ∧ w = y → z x w y φ