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Proof step 5 from proof unequal (previous step)

x = a1 + x0, # x = # y0 + 1, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ x = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ x = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# x0 = # y0, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ a1 ' + x0 = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ a1 ' + x0 = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%len-05: ⊦ #(a ' + x) = (# x)+1;%a: ⊦ (x + y) + z = x + y + z;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2