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Proof step 20 from proof rec (previous step)

a ' + x = x0 + a0 ' + y + a0 ' + z, a0 ∈ x, a ∈ z ∨ a ∈ y ∨ a ∈ x0 ⊦ a ∈ x, ∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z

User applied rule exists right

And obtained the following new goals

Proof step 21:
a ' + x = x0 + a0 ' + y + a0 ' + z, a0 ∈ x, a ∈ z ∨ a ∈ y ∨ a ∈ x0 ⊦ x = x0.tail + a0 ' + y + a0 ' + z, a ∈ x, ∃ a0, x0, y, z. x = x0 + a0 ' + y + a0 ' + z

Rule argument

pos: right 2 terms: a0, x0.tail, y, z abort: F discard: F computed: F

Additional information

Used terms: a0, x0.tail, y, z