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Proof step 22 from proof del-04 (previous step)

#oc(a, y) = m0 + 1, #oc(a, x) = m + 1, a ∈ y, a ∈ x, perm(y, x), ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), a ∈ x -1l a ∨ a ∈ y -1l a ⊦ m = m0

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a abort: F discard: F computed: T

Additional information

Used simplifier rules: 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