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Proof step 21 from proof subset-size (previous step)

a ∈ st0, st1 ⊆ st0 -- a, # st0 ≤ # st1, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ 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 = ∅;%Size-empty: ⊦ # ∅ = 0;%empty-01: ⊦ ∅ -- a = ∅;%subset-empty-01: ⊦ st ⊆ ∅ ↔ st = ∅;%In-empty: ⊦ ¬ a ∈ ∅;%pred-03: ⊦ n -1 = n - 1; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lez: ⊦ 0 ≤ m; from specification nat%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: st0 -- a