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Proof step 4 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 5:
y ≠ [] ⊦ a ' + x = y + z ↔ a = y.first ∧ x = y.rest + z
Proof step 6:
x0 ≠ [], a = x0.first ∧ b ' + y + c ' + z = x0.rest + a ' + c ' + z0 ⊦ a = c, a = b, b = c, dups(z), c ∈ z, a ∈ z, a ∈ y

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] → [a ' + c ' + z0, x0, b ' + y + c ' + z, a] flags: rewrite

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list Used terms: a ' + c ' + z0, x0, b ' + y + c ' + z, a