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Proof step 17 from proof put-firstn-restn (previous step)

z0 = a + y1, a0 ' + y0 + z = y + z0, # y = # y0, a0 ∈ y ∨ a0 ∈ z0 ⊦ z0 = [], y1 = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a0 ' + y0 + z = y + a ' + y1, # y = # y0, a0 ∈ y ∨ a0 = a ∨ a0 ∈ y1 ⊦ y1 = z

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list