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Proof step 2 from proof subset-size (previous step)

∀ bag1. bag ⊆ bag1 → # bag ≤ # bag1 ⊦ bag ++ a ⊆ bag1 → #(bag ++ a) ≤ # bag1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
bag ⊆ bag0, # bag0 < # bag, ∀ bag1. bag ⊆ bag1 → # bag ≤ # bag1 ⊦ # bag = 0

Rule argument

Additional information

Used simplifier rules: subset-plus: ⊦ ¬ ¬ bag0 ++ a ⊆ bag1 ↔ (∃ bag2. bag1 = bag2 ++ a ∧ bag0 ⊆ bag2);%size-rec: ⊦ #(bag ++ a) = # bag + 1;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat