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Proof step 4 from proof delpos-in (previous step)

a ∈ (a0 ' + x0) -1l n, ∀ a, n. a ∈ x0 -1l n → a ∈ x0 ⊦ a = a0, a ∈ x0

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
n = n0 +1, a ∈ (a0 ' + x0) -1l n, ∀ a, n. a ∈ x0 -1l n → a ∈ x0 ⊦ a = a0, a ∈ x0

Rule argument

var: n

Additional information

Used simplifier rules: delpos-base: ⊦ (a ' + x) -1l 0 = x;%nat freely generated by 0, +1;;%a: ⊦ (x + y) + z = x + y + z; from specification list