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Proof step 11 from proof range-eq-append (previous step)

∀ ar0, n0, ar1, n1, m1. range=(ar0, n0, ar1, n1, m + m1) ↔ range=(ar0, n0, ar1, n1, m) ∧ range=(ar0, n0 + m, ar1, n1 + m, m1) ⊦ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m1 + m) ↔ (ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m)) ∧ range=(ar0, n0 + m + 1, ar1, n1 + m + 1, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
ar0[n0] = ar1[n1], ∀ ar0, n0, ar1, n1, m1. range=(ar0, n0, ar1, n1, m + m1) ↔ range=(ar0, n0, ar1, n1, m) ∧ range=(ar0, n0 + m, ar1, n1 + m, m1) ⊦ range=(ar0, n0 + 1, ar1, n1 + 1, m1 + m) ↔ range=(ar0, n0 + 1, ar1, n1 + 1, m) ∧ range=(ar0, n0 + m + 1, ar1, n1 + m + 1, m1)

Proof step 11 from proof range-eq-append (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information