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Proof step 6 from proof range-eq-ext (previous step)

∀ ar0, n0, ar1, n1. range=(ar0, n0, ar1, n1, m0) ↔ n0 + m0 ≤ # ar0 ∧ n1 + m0 ≤ # ar1 ∧ (∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n]) ⊦ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

User applied rule all left

And obtained the following new goals

Proof step 7:
range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ (n0 + 1) + m0 ≤ # ar0 ∧ (n1 + 1) + m0 ≤ # ar1 ∧ (∀ n. n < m0 → ar0[(n0 + 1) + n] = ar1[(n1 + 1) + n]) ⊦ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m0) ↔ n0 + m0 < # ar0 ∧ n1 + m0 < # ar1 ∧ (∀ n. ¬ m0 < n → ar0[n0 + n] = ar1[n1 + n])

Rule argument

pos: left 1 terms: ar0, n0 + 1, ar1, n1 + 1 abort: F discard: T computed: T

Additional information

Used terms: ar0, n0 + 1, ar1, n1 + 1