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Proof step 7 from proof diffone-nodupsget (previous step)

n < # x0, a ∈ x0, ∀ n0, a1. n0 + 1 < # x0 ∧ a1 ∈ x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ (a0 ' + (x0 \ a '))[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), x0 = [], dups(x0), a0 ∈ x0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
n < # x0, pos(a, x0) < n, a ∈ x0, ∀ n0, a1. n0 + 1 < # x0 ∧ a1 ∈ x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ (a0 ' + (x0 \ a '))[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), x0 = [], dups(x0), a0 ∈ x0
Proof step 15:
n < # x0, a ∈ x0, ∀ n0, a1. n0 + 1 < # x0 ∧ a1 ∈ x0 → (x0 \ a1 ')[n0] = (n0 < pos(a1, x0) ⊃ x0[n0];x0[n0 + 1]) ⊦ (a0 ' + (x0 \ a '))[n] = (¬ pos(a, x0) < n ⊃ (a0 ' + x0)[n];x0[n]), x0 = [], pos(a, x0) < n, dups(x0), a0 ∈ x0

Rule argument

fma: pos(a, x0) < n

Additional information

Used formulas: pos(a, x0) < n