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Proof step 10 from proof eq (previous step)

b ' + y + c ' + z = y0 + c ' + a ' + z0, b ∈ y0 ⊦ a = c, a = b, dups(z), c ∈ z, a ∈ z, a ∈ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
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 ' + a ' + z0, y0, y + c ' + z, b] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ a ∈ y ∧ ¬ a ∈ z ∧ a ≠ c → y + c ' + z ≠ y1 + a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ [] + x = x; (s)% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)% ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0); (ws) Used terms: c ' + a ' + z0, y0, y + c ' + z, b