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Proof step 15 from proof restrict-union (previous step)

@(m, m0, m1).lnum = n ∧ @(m, m0, m1) ∈ as0 ∨ @(m, m0, m1) ∈ as1 | n ⊦ m = n ∧ (@(n, m0, m1) ∈ as0 ∨ @(n, m0, m1) ∈ as1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
m = n ∧ @(n, m0, m1) ∈ as0 ∨ @(m, m0, m1) ∈ as1 | n ⊦ m = n ∧ (@(n, m0, m1) ∈ as0 ∨ @(n, m0, m1) ∈ as1)

Rule argument

Additional information

Used simplifier rules: ⊦ @(m, m0, m1).lnum = m; (s)