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Proof step 5 from proof ex-oddeven (previous step)

m + m = n0 ∨ m + m + 1 = n0 ⊦ ∃ m0. m0 + m0 = n0 + 1 ∨ m0 + m0 = n0

User applied rule case distinction

And obtained the following new goals

Proof step 6:
m + m + 1 = n0 ⊦ ∃ m0. m0 + m0 = n0 + 1 ∨ m0 + m0 = n0

Rule argument

pos: left 1

Additional information

Used simplifier rules: even-06: ⊦ m + m ≠ n + n + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%even: ⊦ n + n = m + m ↔ n = m;