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Proof step 3 from proof eq-objs (previous step)

((r => o1 * P1))(H), ((r => o0 * P0))(H) ⊦ o0 = o1

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ ((r => o0 * P))(H) → H[r] = o0
Proof step 5:
((r => o0 * P0))(H), ((r => o1 * P1))(H), ((r => o0 * P0))(H) → H[r] = o0 ⊦ o0 = o1

Rule argument

spec: <toplevel> name: at-rw seq: ⊦ ((r => o0 * P))(H) → H[r] = o0 subst: [r, o0, P, H] → [r, o0, P0, H]

Additional information

Used terms: r, o0, P0, H