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Proof step 21 from proof append-inj (previous step)

# ar0 = # ar1, # ar = # ar2, n < # ar2, ∀ n. n < # ar1 + # ar2 → (ar0 + ar)[n] = (ar1 + ar2)[n] ⊦ ar[n] = ar2[n]

User applied rule all left

And obtained the following new goals

Rule argument

pos: left 4 terms: # ar0 + n abort: F discard: F computed: F

Additional information

Used simplifier rules: append-atp2: n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s) Used terms: # ar0 + n