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Proof step 7 from proof inapp (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
y ≠ [] ⊦ a ' + x = y + z ↔ a = y.first ∧ x = y.rest + z

Rule argument

spec: list inst: name: firsteq seq: y ≠ [] ⊦ a ' + x = y + z ↔ a = y.first ∧ x = y.rest + z subst: [z, y, x, a] → [c ' + c ' + y0, y, x, a] flags: rewrite

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%in-08: ⊦ x.first ∈ x ↔ x ≠ []; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%Nil: ⊦ [] + x = x; from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list Used terms: c ' + c ' + y0, y, x, a