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Proof step 11 from proof subset-plusminus (previous step)

count(a, bag1) ≤ count(a, bag0), ∀ a0. count(a0, bag0) ≤ count(a0, bag1) - (a0 = a ⊃ 1;0) ⊦ count(a, bag1) = 0

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a abort: F discard: F computed: T

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); 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: a