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Proof step 4 from proof inapp (previous step)

(a0 ' + x0) + a ' + y = z + a ' + y0, a0 ∈ z, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ y ∨ a ∈ x0 ∨ y = y0 ∧ x0 = z ⊦ a = a0, a ∈ x0, a ∈ y, y = y0 ∧ a0 ' + x0 = z

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
y ≠ [] ⊦ a ' + x = y + z ↔ a = y.first ∧ x = y.rest + z
Proof step 6:
z + a ' ≠ [], a0 = (z + a ').first ∧ x0 + a ' + y = (z + a ').rest + y0, a0 ∈ z, ∀ a, y, z, y0. x0 + a ' + y = z + a ' + y0 → a ∈ y ∨ a ∈ x0 ∨ y = y0 ∧ x0 = z ⊦ a = a0, a ∈ x0, a ∈ y, y = y0 ∧ a0 ' + x0 = 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] → [y0, z + a ', x0 + a ' + y, a0] flags: rewrite, current

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = [];%e: ⊦ a ' ≠ [];%a: ⊦ (x + y) + z = x + y + z; Used terms: y0, z + a ', x0 + a ' + y, a0