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Proof step 24 from proof key-size (previous step)

# st0 = m + 1 ∧ # st0 ≠ 0, a ∈ st0, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0, (a0 ∈ st1 ↔ a0 ∈ st0 -- a) → # st1 = m ⊦ # st1 + 1 = # st0, a ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
# 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

Rule argument

Additional information

Used simplifier rules: empty: ⊦ # st = 0 ↔ st = ∅;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2