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Proof step 15 from proof put-firstn-restn (previous step)

z0 ≠ [], a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ a ' + z0.rest = a ' + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ z0 = [], z0.rest = z

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list