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Proof step 8 from proof longest-common-prefix (previous step)

∀ x. ∃ z, x0, y1. x = z + x0 ∧ y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.first ≠ y1.first) ⊦ ∃ z, x, y. a0 ' + x1 = z + x ∧ a ' + y0 = z + y ∧ (x ≠ [] ∧ y ≠ [] → x.first ≠ y.first)

User applied rule cut formula

And obtained the following new goals

Proof step 9:
a = a0, ∀ x. ∃ z, x0, y1. x = z + x0 ∧ y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.first ≠ y1.first) ⊦ ∃ z, x, y. a0 ' + x1 = z + x ∧ a ' + y0 = z + y ∧ (x ≠ [] ∧ y ≠ [] → x.first ≠ y.first)
Proof step 13:
∀ x. ∃ z, x0, y1. x = z + x0 ∧ y0 = z + y1 ∧ (x0 ≠ [] ∧ y1 ≠ [] → x0.first ≠ y1.first) ⊦ a = a0, ∃ z, x, y. a0 ' + x1 = z + x ∧ a ' + y0 = z + y ∧ (x ≠ [] ∧ y ≠ [] → x.first ≠ y.first)

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0