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

a ' + x = y + c ' + a0 ' + y0, c ∈ y0, a ∈ y0 ∨ a ∈ y, a ∈ y ∨ a ∈ y0 ⊦ a = a0, a = a1, c = a1, dups(x), a1 ∈ x, a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

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 ' + a0 ' + y0, y, x, a] flags: rewrite

Additional information

Used simplifier rules: in-03-01: ⊦ b ∈ y → dups(x + b ' + a ' + y);%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%in-empty: ⊦ ¬ a ∈ []; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-12: ⊦ a ∈ y → a ∈ x + y; from specification list Used terms: c ' + a0 ' + y0, y, x, a