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Proof step 16 from proof diff-twice-union (previous step)

∀ y, z. (x0 \ y) \ z = x0 \ rmdup(y + z) ⊦ a ' + ((x0 \ y) \ z) = (a ' + x0) \ rmdup(y + z), a ∈ z, a ∈ y

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 17:
∀ y, z. (x0 \ y) \ z = x0 \ rmdup(y + z) ⊦ a ∈ rmdup(y + z), a ' + ((x0 \ y) \ z) = (a ' + x0) \ rmdup(y + z), a ∈ z, a ∈ y

Rule argument

fma: a ∈ rmdup(y + z)

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%a: ⊦ (x + y) + z = x + y + z; from specification list%rm-in: ⊦ a ∈ rmdup(x + y) ↔ a ∈ x ∨ a ∈ y; from specification list-dup Used formulas: a ∈ rmdup(y + z)