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Proof step 4 from proof bag-rm-add-eq (previous step)

∀ a0. count(a0, (bag -- a) ++ a) = count(a0, bag) ⊦ a ∈ bag

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a abort: F discard: T computed: T

Additional information

Used simplifier rules: count-del: ⊦ count(a, bag -- b) = count(a, bag) - (a = b ∧ count(a, bag) ≠ 0 ⊃ 1;0);%in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; from specification nat%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a