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Proof step 6 from proof blast-first (previous step)

a0 ' + y = y0 + a1 ' ∧ a0 ' + y ≠ [], a ∈ y0, a = a0 ∨ a ∈ y ⊦ y0.first = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 ' + y = y0 + a1 ', a ∈ y0, a = a0 ∨ a ∈ y, a0 ∈ y0 ∨ a0 = a1, a1 = a0 ∨ a1 ∈ y ⊦ y0.first = a0

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list