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Proof step 6 from proof subsize-eq (previous step)

# s = # s0, ∀ a. a ∈ s → a ∈ s0, a ∈ s0 ⊦ a ∈ s

User applied rule all left

And obtained the following new goals

Proof step 7:
# s = # s0, a ∈ s → a ∈ s0, ∀ a. a ∈ s → a ∈ s0, a ∈ s0 ⊦ a ∈ s

Rule argument

pos: left 2 terms: a abort: F discard: F computed: T

Additional information

Used terms: a