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Proof step 5 from proof del-05 (previous step)

a ∈ x, perm(x -1l a, x), ∀ a0. a0 ∈ x ∨ a0 ∈ x -1l a → #oc(a0, x) = #oc(a0, x -1l a) ⊦

User applied rule all left

And obtained the following new goals

Proof step 6:
a ∈ x, perm(x -1l a, x), a ∈ x ∨ a ∈ x -1l a → #oc(a, x) = #oc(a, x -1l a), ∀ a0. a0 ∈ x ∨ a0 ∈ x -1l a → #oc(a0, x) = #oc(a0, x -1l a) ⊦

Rule argument

pos: left 3 terms: a abort: F discard: F computed: T

Additional information

Used terms: a