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Proof step 1 from proof map-putex (conclusion)

r ∈ H ⊦ ((r => o0 * P))(H[r, o1]) ↔ o0 = o1 ∧ ((r =>_ * P))(H)

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((r => o0 * P))(H[r, o1]), r ∈ H ⊦ o0 = o1 ∧ ((r =>_ * P))(H)
Proof step 12:
o0 = o1, ((r =>_ * P))(H), r ∈ H ⊦ ((r => o0 * P))(H[r, o1])

Rule argument

pos: right 1

Additional information