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Proof step 18 from proof key-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0 ⊦ # st1 + 1 = # st0, a ∈ st1

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ st → # st[a, d] = # st + 1;