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Proof step 9 from proof last-ex (previous step)

⊦ a = a0, ∃ y0, z0. (y + a ' + z) + a0 ' = y0 + a ' + z0 ∧ z0 ≠ []

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: y, z + a0 ' abort: F discard: T computed: F

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s) Used terms: y, z + a0 '