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Proof step 8 from proof inapp (previous step)

a ' + x = y + c ' + c ' + y0, a ∈ y, c = a ∨ c ∈ x ⊦ a = b, dups(x), a ∈ x, b ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
y ≠ [] ⊦ a ' + x = y + z ↔ a = y.head ∧ x = y.tail + z

Rule argument

spec: list name: firsteq seq: y ≠ [] ⊦ a ' + x = y + z ↔ a = y.head ∧ x = y.tail + z subst: [z, y, x, a] → [c ' + c ' + y0, y, x, a] flags: <all paths>

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);% ⊦ x.head ∈ x ↔ x ≠ []; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ [] + x = x; (s)% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws) Used terms: c ' + c ' + y0, y, x, a