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Proof step 5 from proof eq-01 (previous step)

a ' + b ' + y + c ' + z = x0 + a ' + c ' + z0 ⊦ a = c, a = b, b = c, dups(z), c ∈ z, a ∈ z, a ∈ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
y ≠ [] ⊦ a ' + x = y + z ↔ a = y.head ∧ x = y.tail + z
Proof step 7:
b ' + y + c ' + z = x0.tail + a ' + c ' + z0, a = x0.head ⊦ a = b, a = c, b = c, x0 = [], dups(z), a ∈ y, a ∈ z, c ∈ 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] → [a ' + c ' + z0, x0, b ' + y + c ' + z, a] flags: <all paths>

Additional information

Used simplifier rules: ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws) Used terms: a ' + c ' + z0, x0, b ' + y + c ' + z, a