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Proof step 3 from proof len-06 (previous step)

# y = # y0, x + y = z + y0 ⊦ x = z ∧ y = y0

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0)
Proof step 5:
# y = # y0, x + y = z + y0, # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0) ⊦ x = z ∧ y = y0

Rule argument

spec: <toplevel> name: len-01 seq: ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0) subst: [x, z, y, y0] → [x, z, y, y0]

Additional information

Used terms: x, z, y, y0