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Proof step 4 from proof dups-firstoc-rew (previous step)

a ∈ x0 ∨ dups(x0), dups(x0) → x0 = y + a0 ' + z ∧ a0 ∈ y ∧ ¬ dups(y) ⊦ ∃ y0, a1, z0. a ' + x0 = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

User applied rule cut formula

And obtained the following new goals

Proof step 5:
dups(x0), a ∈ x0 ∨ dups(x0), dups(x0) → x0 = y + a0 ' + z ∧ a0 ∈ y ∧ ¬ dups(y) ⊦ ∃ y0, a1, z0. a ' + x0 = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)
Proof step 12:
a ∈ x0 ∨ dups(x0), dups(x0) → x0 = y + a0 ' + z ∧ a0 ∈ y ∧ ¬ dups(y) ⊦ dups(x0), ∃ y0, a1, z0. a ' + x0 = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

fma: dups(x0)

Additional information

Used formulas: dups(x0)