[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 5 from proof append-at (previous step)

n < # ar0 + # ar1 ⊦ (ar0 + ar1)[n] = (n < # ar0 ⊃ ar0[n];ar1[n - # ar0]), # ar0 + # ar1 = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 1:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]

Rule argument

pos: right 1

Additional information

[Concretesimprules((Csimpnamedseq(append-at1,,, ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]) Csimpnamedseq(append-a,,, ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1) Csimpnamedseq(append-at2,,, ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]) Csimpnamedseq(append-a,,, ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1) Csimpnamedseq(nat,,lels, ⊦ ¬ m < n ↔ n ≤ m) Csimpnamedseq(nat-basic2,,a, ⊦ (m + n0) + n1 = m + n0 + n1) Csimpnamedseq(nat-basic2,,c, ⊦ m + n = n + m)))]