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Proof step 6 from proof inapp-01 (previous step)

a ' + x = y + c ' + a0 ' + y0, c ∈ y0, a0 ∈ x, c ∈ x ∨ c = a, a ∈ y ∨ a ∈ y0, c = a ∨ c ∈ x ⊦ a = b, c = b, dups(x), a ∈ x, b ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

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 ' + a0 ' + y0, y, x, a] flags: <all paths>

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;%in-03-01: ⊦ b ∈ y → dups(x + b ' + a ' + y);% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)% ⊦ [] + x = x; (s)% ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; (ws)% ⊦ ¬ a ∈ []; (s)%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss) Used terms: c ' + a0 ' + y0, y, x, a