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Proof step 16 from proof last-elim (previous step)

a1 ' + x = y1 + a ', a0 = a3, ∀ a2, y0. a2 = (a1 ' + x).last ∧ y0 = (a1 ' + x).butlast ↔ a1 ' + x = y0 + a2 ' ⊦ a = (a1 ' + x).last ∧ a3 = a0 ∧ y1 = (a1 ' + x).butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
a1 ' + x = y1 + a ', ∀ a2, y0. a2 = (y1 + a ').last ∧ y0 = (y1 + a ').butlast ↔ y1 + a ' = y0 + a2 ', a1 ∈ y1 ∨ a1 = a, a = a1 ∨ a ∈ x ⊦ a = (y1 + a ').last ∧ y1 = (y1 + a ').butlast

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' + x ↔ a = b ∨ a ∈ x; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list