Metamath Proof Explorer


Theorem ax13fromc9

Description: Derive ax-13 from ax-c9 and other older axioms.

This proof uses newer axioms ax-4 and ax-6 , but since these are proved from the older axioms above, this is acceptable and lets us avoid having to reprove several earlier theorems to use ax-c4 and ax-c10 . (Contributed by NM, 21-Dec-2015) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion ax13fromc9 ( ¬ 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → ∀ 𝑥 𝑦 = 𝑧 ) )

Proof

Step Hyp Ref Expression
1 ax-c5 ( ∀ 𝑥 𝑥 = 𝑦𝑥 = 𝑦 )
2 1 con3i ( ¬ 𝑥 = 𝑦 → ¬ ∀ 𝑥 𝑥 = 𝑦 )
3 ax-c5 ( ∀ 𝑥 𝑥 = 𝑧𝑥 = 𝑧 )
4 3 con3i ( ¬ 𝑥 = 𝑧 → ¬ ∀ 𝑥 𝑥 = 𝑧 )
5 ax-c9 ( ¬ ∀ 𝑥 𝑥 = 𝑦 → ( ¬ ∀ 𝑥 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → ∀ 𝑥 𝑦 = 𝑧 ) ) )
6 2 4 5 syl2im ( ¬ 𝑥 = 𝑦 → ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → ∀ 𝑥 𝑦 = 𝑧 ) ) )
7 ax13b ( ( ¬ 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → ∀ 𝑥 𝑦 = 𝑧 ) ) ↔ ( ¬ 𝑥 = 𝑦 → ( ¬ 𝑥 = 𝑧 → ( 𝑦 = 𝑧 → ∀ 𝑥 𝑦 = 𝑧 ) ) ) )
8 6 7 mpbir ( ¬ 𝑥 = 𝑦 → ( 𝑦 = 𝑧 → ∀ 𝑥 𝑦 = 𝑧 ) )