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Proof step 23 from proof key-size (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
# 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

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# st0, m] flags:

Additional information

Used simplifier rules: empty: ⊦ # st = 0 ↔ st = ∅;%Size-empty: ⊦ # ∅ = 0;%In-empty: ⊦ ¬ a ∈ ∅;%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # st0, m