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Proof step 8 from proof key-size (previous step)

a ∈ st1, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0 ⊦ # st1 = # st0

User applied rule all left

And obtained the following new goals

Proof step 9:
a ∈ st1, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0, a ∈ st1[a, d] ↔ a ∈ st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0 ⊦ # st1 = # st0

Rule argument

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

Additional information

Used terms: a