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Proof step 2 from proof otwo-02 (previous step)

2 * m0 = m + 1 ⊦ ∃ m1. m = 2 * m1 + 1

User applied rule exists right

And obtained the following new goals

Proof step 3:
2 * m0 = m + 1 ⊦ m = 2 * (m0 - 1) + 1, ∃ m1. m = 2 * m1 + 1

Rule argument

pos: right 1 terms: m0 - 1 abort: F discard: F computed: F

Additional information

Used terms: m0 - 1