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Proof step 16 from proof restrict-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ adr ∈ as | n ↔ adr.lnum = n ∧ adr ∈ as
Proof step 18:
m = n ∧ @(n, m0, m1) ∈ as0 ∨ @(m, m0, m1).lnum = n ∧ @(m, m0, m1) ∈ as1 ⊦ m = n ∧ (@(n, m0, m1) ∈ as0 ∨ @(n, m0, m1) ∈ as1)

Rule argument

spec: <toplevel> name: restrict-leb seq: ⊦ adr ∈ as | n ↔ adr.lnum = n ∧ adr ∈ as subst: [n, as, adr] → [n, as1, @(m, m0, m1)] flags: <all paths>

Additional information

Used terms: n, as1, @(m, m0, m1)