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

a ∈ (a0 + x0) -p n, ∀ a, n. a ∈ x0 -p n → a ∈ x0 ⊦ a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)% ⊦ a + x = a ' + x; (s)