[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
# st0 = m + 1, a ∈ st0, (a0 ∈ st1 ↔ a0 ∈ st0 -- a) → # st1 = m, ∀ a0. a0 ∈ st1[a, d] ↔ a0 ∈ st0, ∀ st0. (∀ a. a ∈ st1 ↔ a ∈ st0) → # st1 = # st0 ⊦ # st0 = 0, # st1 + 1 = # st0, a ∈ st1

Rule argument

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

Additional information

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