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Proof step 9 from proof ext-rec (previous step)

# ar0 = # ar1, n0 < # ar1, ∀ m. m < # ar1 → ar1[m] = (n = m ⊃ a;ar0[m]) ⊦ ar1[n0] = ar0[n, a][n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m ≠ n → ar[n, a][m] = ar[m]
Proof step 11:
# ar0 = # ar1, n0 = n, n0 < # ar1, ∀ m. m < # ar1 → ar1[m] = (n = m ⊃ a;ar0[m]) ⊦ ar1[n0] = ar0[n, a][n0]
Proof step 13:
# ar0 = # ar1, n0 < # ar1, ∀ m. m < # ar1 → ar1[m] = (n = m ⊃ a;ar0[m]) ⊦ ar1[n0] = ar0[n0], n0 = n

Rule argument

spec: <toplevel> name: Put-other seq: ⊦ m ≠ n → ar[n, a][m] = ar[m] subst: [m, a, n, ar] → [n0, a, n, ar0] flags: <all paths>

Additional information

Used terms: n0, a, n, ar0