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Proof step 21 from proof key-size (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 22:
a ∈ st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0, (∀ a0. a0 ∈ st1 ↔ a0 ∈ st0 -- a) → # st1 = #(st0 -- a), ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0 ⊦ # st1 + 1 = # st0, a ∈ st1

Rule argument

pos: left 3 terms: st0 -- a abort: F discard: F computed: F

Additional information

Used terms: st0 -- a