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Proof step 8 from proof eq-bothsubset (previous step)

∀ x. S(x) → S0(x), ∀ x. S0(x) → S(x) ⊦ S(x) ↔ S0(x)

User applied rule case distinction

And obtained the following new goals

Proof step 9:
S(x), ∀ x. S(x) → S0(x), ∀ x. S0(x) → S(x) ⊦ S0(x)
Proof step 10:
S0(x), ∀ x. S(x) → S0(x), ∀ x. S0(x) → S(x) ⊦ S(x)

Rule argument

pos: right 1

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