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Proof step 2 from proof to-ards-no-dups (previous step)

∀ n. dups(lax0.to-adr(n)) ↔ dups(lax0) ⊦ dups((ladr + lax0).to-adr(n)) ↔ dups(ladr + lax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ 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)

Rule argument

Additional information

Used simplifier rules: adrs-rec: ⊦ (ladr ' + lax).to-adr(n) = @(n, ladr.off, ladr.size) + lax.to-adr(n);% ⊦ dups(ladr ' + lax) ↔ ladr ∈ lax ∨ dups(lax); (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ dups(adr ' + ax) ↔ adr ∈ ax ∨ dups(ax); (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ adr + ax = adr ' + ax; (s)% ⊦ ladr + lax = ladr ' + lax; (s)