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Proof step 31 from proof diff-union (previous step)

a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ (a ∈ x0 ∨ a ∈ y ⊃ rmdup(x0 + y);a ' + rmdup(x0 + y)) \ z = rmdup((x0 \ z) + (y \ z))

User applied rule case distinction

And obtained the following new goals

Proof step 32:
a ∈ x0 ∨ a ∈ y, a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ rmdup(x0 + y) \ z = rmdup((x0 \ z) + (y \ z))
Proof step 33:
¬(a ∈ x0 ∨ a ∈ y), a ∈ z, ∀ y0, z0. rmdup(x0 + y0) \ z0 = rmdup((x0 \ z0) + (y0 \ z0)) ⊦ (a ' + rmdup(x0 + y)) \ z = rmdup((x0 \ z) + (y \ z))

Rule argument

pos: right 1

Proof step 31 from proof diff-union (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information