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Proof step 7 from proof prefixpoint-aux2 (previous step)

f(S) ⊆ S, S0(x), ∀ S. f(S) ⊆ S → S0 ⊆ S ⊦ S(x)

User applied rule all left

And obtained the following new goals

Proof step 8:
f(S) ⊆ S, S0(x), f(S) ⊆ S → S0 ⊆ S ⊦ S(x)

Rule argument

pos: left 3 terms: S abort: F discard: T computed: T

Additional information

Used terms: S