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Proof step 22 from proof subset-size (previous step)

a ∈ st0, # st0 ≤ # st1, st1 ⊆ st0 -- a, ∀ st. ¬ # st < # st1 ∨ ¬ st1 ⊆ st ⊦ st1[a, st0[a]] = ∅, a ∈ st1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: st0 -- a

Additional information

Used simplifier rules: size: ⊦ a ∈ st → #(st -- a) = (# st)-1;%empty: ⊦ # st = 0 ↔ st = ∅;%Del-empty: ⊦ ∅ -- a = ∅;%subset-empty-01: ⊦ st ⊆ ∅ ↔ st = ∅;%In-empty: ⊦ ¬ a ∈ ∅;% ⊦ n -1 = n - 1; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: st0 -- a