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Proof step 25 from proof key-size (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 26:
# 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

Rule argument

pos: left 4 terms: a0 abort: F discard: F computed: F

Additional information

Used terms: a0