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Proof step 3 from proof to-ards-no-dups (previous step)

∀ n. dups(lax0.to-adr(n)) ↔ dups(lax0) ⊦ @(n, ladr.off, ladr.size) ∈ lax0.to-adr(n) ∨ dups(lax0.to-adr(n)) ↔ ladr ∈ lax0 ∨ dups(lax0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
ladr = @(m, m0), ∀ n. dups(lax0.to-adr(n)) ↔ dups(lax0) ⊦ @(n, m, m0) ∈ lax0.to-adr(n) ∨ dups(lax0.to-adr(n)) ↔ ladr ∈ lax0 ∨ dups(lax0)

Rule argument

spec: laddress name: elim seq: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0) subst: [ladr, m, m0] → [ladr, m, m0]

Additional information

Used simplifier rules: ⊦ m = ladr.off ∧ m0 = ladr.size ↔ ladr = @(m, m0); (elim) Used terms: ladr, m, m0