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Proof step 10 from proof key-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st
Proof step 12:
a ∈ st1, a = a ∨ a ∈ st1, a ∈ st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0 ⊦ # st1 = # st0

Rule argument

spec: inst: name: In-insert seq: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st subst: [d, b, st, a] → [d, a, st1, a] flags: rewrite, current

Additional information

Used terms: d, a, st1, a