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Proof step 5 from proof to-ards-no-dups (previous step)

∀ n. dups(lax0.to-adr(n)) ↔ dups(lax0) ⊦ @(n, m, m0) ∈ lax0.to-adr(n) ∨ dups(lax0.to-adr(n)) ↔ @(m, m0) ∈ lax0 ∨ dups(lax0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
dups(lax0), ∀ n. dups(lax0.to-adr(n)) ↔ dups(lax0) ⊦ @(n, m, m0) ∈ lax0.to-adr(n) ∨ dups(lax0.to-adr(n)) ↔ @(m, m0) ∈ lax0 ∨ dups(lax0)
Proof step 8:
∀ n. dups(lax0.to-adr(n)) ↔ dups(lax0) ⊦ @(n, m, m0) ∈ lax0.to-adr(n) ∨ dups(lax0.to-adr(n)) ↔ @(m, m0) ∈ lax0 ∨ dups(lax0), dups(lax0)

Rule argument

fma: dups(lax0)

Additional information

Used formulas: dups(lax0)