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Proof step 18 from proof put-firstn-restn (previous step)

a0 ' + y0 + z = y + a ' + y1, # y = # y0, a0 ∈ y ∨ a0 = a ∨ a0 ∈ y1 ⊦ y1 = z

User applied rule constructor cut

And obtained the following new goals

Proof step 19:
y = a1 + y2, a0 ' + y0 + z = y + a ' + y1, # y = # y0, a0 ∈ y ∨ a0 = a ∨ a0 ∈ y1 ⊦ y1 = z

Rule argument

var: y

Additional information

Used simplifier rules: list freely generated by [], +;;%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%size-01: ⊦ # [] = 0; from specification list-data%l: ⊦ # x = 0 ↔ x = []; from specification list-data%in-empty: ⊦ ¬ a ∈ []; from specification list