Metamath Proof Explorer

Theorem sbi1OLD

Description: Obsolete version of sbi1 as of 24-Jul-2023. Removal of implication from substitution. (Contributed by NM, 14-May-1993) (New usage is discouraged.) (Proof modification is discouraged.)

		Ref	Expression
	Assertion	sbi1OLD	$⊢ [y/ x] (φ \to ψ) \to ([y/ x] φ \to [y/ x] ψ)$

Proof

Step	Hyp	Ref	Expression
1		sbequ2	$⊢ x = y \to ([y/ x] φ \to φ)$
2		sbequ2	$⊢ x = y \to ([y/ x] (φ \to ψ) \to (φ \to ψ))$
3	1 2	syl5d	$⊢ x = y \to ([y/ x] (φ \to ψ) \to ([y/ x] φ \to ψ))$
4		sbequ1	$⊢ x = y \to (ψ \to [y/ x] ψ)$
5	3 4	syl6d	$⊢ x = y \to ([y/ x] (φ \to ψ) \to ([y/ x] φ \to [y/ x] ψ))$
6	5	sps	$⊢ \forall x x = y \to ([y/ x] (φ \to ψ) \to ([y/ x] φ \to [y/ x] ψ))$
7		sb4OLD	$⊢ \neg \forall x x = y \to ([y/ x] φ \to \forall x (x = y \to φ))$
8		sb4OLD	$⊢ \neg \forall x x = y \to ([y/ x] (φ \to ψ) \to \forall x (x = y \to (φ \to ψ)))$
9		ax-2	$⊢ (x = y \to (φ \to ψ)) \to ((x = y \to φ) \to (x = y \to ψ))$
10	9	al2imi	$⊢ \forall x (x = y \to (φ \to ψ)) \to (\forall x (x = y \to φ) \to \forall x (x = y \to ψ))$
11		sb2	$⊢ \forall x (x = y \to ψ) \to [y/ x] ψ$
12	10 11	syl6	$⊢ \forall x (x = y \to (φ \to ψ)) \to (\forall x (x = y \to φ) \to [y/ x] ψ)$
13	8 12	syl6	$⊢ \neg \forall x x = y \to ([y/ x] (φ \to ψ) \to (\forall x (x = y \to φ) \to [y/ x] ψ))$
14	7 13	syl5d	$⊢ \neg \forall x x = y \to ([y/ x] (φ \to ψ) \to ([y/ x] φ \to [y/ x] ψ))$
15	6 14	pm2.61i	$⊢ [y/ x] (φ \to ψ) \to ([y/ x] φ \to [y/ x] ψ)$