[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 12 from proof challenge3b (previous step)

ar[m] = ar[n0], m0 < # ar, m < # ar, n0 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0] ∨ ar[n0] = a, ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a), ¬ (¬ m0 < m1 → (m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a)) ∨ (∀ m2, n1. ar[m2] ≠ ar[n1] ∨ ar[m2] = a ∨ m2 = n1 ∨ ¬ m2 < # ar ∨ ¬ n1 < # ar) ⊦ ar[n0] = a, m = n0

User applied rule case distinction

And obtained the following new goals

Proof step 10:
¬ (¬ m0 < m1 → (m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a)), ar[m] = ar[n0], m0 < # ar, m < # ar, n0 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0] ∨ ar[n0] = a, ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a) ⊦ ar[n0] = a, m = n0
Proof step 11:
∀ m2, n1. ar[m2] ≠ ar[n1] ∨ ar[m2] = a ∨ m2 = n1 ∨ ¬ m2 < # ar ∨ ¬ n1 < # ar, ar[m] = ar[n0], m0 < # ar, m < # ar, n0 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0] ∨ ar[n0] = a, ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a) ⊦ ar[n0] = a, m = n0

Rule argument

[CASEDARG : pos: left 7, ()]

Additional information

[[CONCRETESIMPRULES : ()]]