Theorem solvedsub-P6b 713

Description: solvedsub-P6a 711 with result substituted back into hypothesis.

Requires the existence of '𝜓₁(𝑥₁)' as a replacement for '𝜓(𝑥)' and '𝜒₁(𝑦₁)' as a replacement for '𝜒(𝑦)'. Also, neither '𝑦' nor '𝑦₁' can occur in '𝜑' and '𝑦' cannot occur in '𝑡'.

Note that trnsvsubw-P6 710 must be used to show that '(𝑥 = 𝑡 → (𝜑 ↔ 𝜒))', and from there we can substitute the explicit formula from solvedsub-P6a 711 for '𝜒'.

Hypotheses

Ref	Expression
solvedsub-P6b.1	⊢ (𝑥 = 𝑥₁ → (𝜓 ↔ 𝜓₁))
solvedsub-P6b.2	⊢ Ⅎ𝑥𝜓
solvedsub-P6b.3	⊢ (𝑦 = 𝑦₁ → (𝜒 ↔ 𝜒₁))
solvedsub-P6b.4	⊢ Ⅎ𝑦𝜒
solvedsub-P6b.5	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
solvedsub-P6b.6	⊢ (𝑦 = 𝑡 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
solvedsub-P6b	⊢ (𝑥 = 𝑡 → (𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))

Distinct variable groups:   𝜓,𝑥₁   𝜓₁,𝑥   𝜒,𝑦₁   𝜒₁,𝑦   𝜑,𝑦,𝑦₁   𝑡,𝑦   𝑥,𝑥₁,𝑦,𝑦₁

Proof of Theorem solvedsub-P6b

Step	Hyp	Ref	Expression
1		solvedsub-P6b.4	. . 3 ⊢ Ⅎ𝑦𝜒
2		solvedsub-P6b.3	. . 3 ⊢ (𝑦 = 𝑦₁ → (𝜒 ↔ 𝜒₁))
3		solvedsub-P6b.5	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4		solvedsub-P6b.6	. . 3 ⊢ (𝑦 = 𝑡 → (𝜓 ↔ 𝜒))
5	1, 2, 3, 4	trnsvsubw-P6 710	. 2 ⊢ (𝑥 = 𝑡 → (𝜑 ↔ 𝜒))
6		solvedsub-P6b.1	. . . 4 ⊢ (𝑥 = 𝑥₁ → (𝜓 ↔ 𝜓₁))
7		solvedsub-P6b.2	. . . 4 ⊢ Ⅎ𝑥𝜓
8	6, 7, 2, 1, 3, 4	solvedsub-P6a 711	. . 3 ⊢ (𝜒 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
9	8	subbir-P3.41b.RC 335	. 2 ⊢ ((𝜑 ↔ 𝜒) ↔ (𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
10	5, 9	subimr2-P4.RC 543	1 ⊢ (𝑥 = 𝑡 → (𝜑 ↔ ∀𝑦(𝑦 = 𝑡 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))

Syntax hints:  term-obj 1   = wff-equals 6  ∀wff-forall 8   → wff-imp 10   ↔ wff-bi 104  Ⅎwff-nfree 681

This theorem was proved from axioms:  ax-L1 11  ax-L2 12  ax-L3 13  ax-MP 14  ax-GEN 15  ax-L4 16  ax-L5 17  ax-L6 18  ax-L7 19

This theorem depends on definitions:  df-bi-D2.1 107  df-and-D2.2 133  df-or-D2.3 145  df-true-D2.4 155  df-rcp-AND3 161  df-exists-D5.1 596  df-nfree-D6.1 682

This theorem is referenced by: (None)