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Proof step 55 from proof challenge3b (previous step)

¬ bv0, ar[m] = ar[n], ar1 = ar, a1 = a, m2 = 0, m ≠ n, ar[n] ≠ a, n < # ar, m < # ar ⊦ ⦉while m2 < # ar1 ∧ ¬ bv0 do { let n1 = m2 + 1 in while n1 < # ar1 ∧ ¬ bv0 do if ar1[m2] = ar1[n1] ∧ ar1[m2] ≠ a1 then m1 := m2, n2 := n1, bv0 := true else n1 := n1 + 1 ; m2 := m2 + 1 }⦊ (ar[m1] = ar[n2] ∧ ar[m1] ≠ a ∧ m1 ≠ n2 ∧ m1 < # ar ∧ n2 < # ar)

User applied rule invariant right

And obtained the following new goals

Proof step 3:
¬ bv0, ar[m] = ar[n], ar1 = ar, a1 = a, m2 = 0, m ≠ n, ar[n] ≠ a, n < # ar, m < # ar, (bv0 ⊃ (ar1[m1] ≠ ar1[n2] ∨ ar[m1] = a1 ∨ m1 = n2 ∨ ¬ m1 < # ar1 ∨ ¬ n2 < # ar1) ; (∃ m1. ¬ (m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1))) ) ∨ ¬ m2 ≤ # ar1 ∨ ar1 ≠ ar ∨ a1 ≠ a ∨ (∀ m, n. ¬ (ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1)) ⊦
Proof step 44:
m2 < # ar1 ∧ ¬ bv0, bv0 ⊃ (ar1[m1] = ar1[n2] ∧ ar[m1] ≠ a1 ∧ m1 ≠ n2 ∧ m1 < # ar1 ∧ n2 < # ar1) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1)), m2 ≤ # ar1, ar1 = ar, a1 = a, ∃ m, n. ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1, m2 = m0 ⊦ ⦉let n1 = m2 + 1 in while n1 < # ar1 ∧ ¬ bv0 do if ar1[m2] = ar1[n1] ∧ ar1[m2] ≠ a1 then m1 := m2, n2 := n1, bv0 := true else n1 := n1 + 1⦊ ⦉m2 := m2 + 1⦊ ( (bv0 ⊃ (ar1[m1] = ar1[n2] ∧ ar[m1] ≠ a1 ∧ m1 ≠ n2 ∧ m1 < # ar1 ∧ n2 < # ar1) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1)) ) ∧ m2 ≤ # ar1 ∧ ar1 = ar ∧ a1 = a ∧ (∃ m, n. ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1) ∧ # ar - m2 < # ar - m0)
Proof step 54:
¬ (m2 < # ar1 ∧ ¬ bv0), bv0 ⊃ (ar1[m1] = ar1[n2] ∧ ar[m1] ≠ a1 ∧ m1 ≠ n2 ∧ m1 < # ar1 ∧ n2 < # ar1) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1)), m2 ≤ # ar1, ar1 = ar, a1 = a, ∃ m, n. ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1, ar[m1] ≠ ar[n2] ∨ ar[m1] = a ∨ m1 = n2 ∨ ¬ m1 < # ar ∨ ¬ n2 < # ar ⊦

Rule argument

[invariant-arg: # ar - m2 , ( (bv0 ⊃ ( ar1[m1] = ar1[n2] ∧ ar[m1] ≠ a1 ∧ m1 ≠ n2 ∧ m1 < # ar1 ∧ n2 < # ar1) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1)) ) ∧ m2 ≤ # ar1 ∧ ar1 = ar ∧ a1 = a ∧ (∃ m, n. ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1)) , pos: right 1 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (( (bv0 ⊃ ( ar1[m1] = ar1[n2] ∧ ar[m1] ≠ a1 ∧ m1 ≠ n2 ∧ m1 < # ar1 ∧ n2 < # ar1) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar1 → ar1[m1] ≠ ar1[n] ∨ ar[m1] = a1)) ) ∧ m2 ≤ # ar1 ∧ ar1 = ar ∧ a1 = a ∧ (∃ m, n. ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1)), # ar - m2 = # ar - m2)]]