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Proof step 26 from proof key-size (previous step)

# st0 = m + 1, a ∈ st0, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0, a0 ∈ st1[a, d] ↔ a0 ∈ st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0 ⊦ a0 ∈ st1 ↔ a0 ∈ st0 -- a, st0 = ∅, # st1 = m, a ∈ st1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;